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Popular Trigonometry >

cos^3(x)= 1/(4(3cos(x)+cos(3x)))

  • Pre Algebra
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Solution

cos3(x)=4(3cos(x)+cos(3x))1​

Solution

x=0.88929…+2πn,x=2π−0.88929…+2πn,x=2.25229…+2πn,x=−2.25229…+2πn
+1
Degrees
x=50.95278…∘+360∘n,x=309.04721…∘+360∘n,x=129.04721…∘+360∘n,x=−129.04721…∘+360∘n
Solution steps
cos3(x)=4(3cos(x)+cos(3x))1​
Subtract 4(3cos(x)+cos(3x))1​ from both sidescos3(x)−4(3cos(x)+cos(3x))1​=0
Simplify cos3(x)−4(3cos(x)+cos(3x))1​:4(3cos(x)+cos(3x))4cos3(x)(3cos(x)+cos(3x))−1​
cos3(x)−4(3cos(x)+cos(3x))1​
Convert element to fraction: cos3(x)=4(3cos(x)+cos(3x))cos3(x)4(3cos(x)+cos(3x))​=4(3cos(x)+cos(3x))cos3(x)⋅4(3cos(x)+cos(3x))​−4(3cos(x)+cos(3x))1​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=4(3cos(x)+cos(3x))cos3(x)⋅4(3cos(x)+cos(3x))−1​
4(3cos(x)+cos(3x))4cos3(x)(3cos(x)+cos(3x))−1​=0
g(x)f(x)​=0⇒f(x)=04cos3(x)(3cos(x)+cos(3x))−1=0
Rewrite using trig identities
−1+(cos(3x)+3cos(x))⋅4cos3(x)
cos(3x)=4cos3(x)−3cos(x)
cos(3x)
Rewrite using trig identities
cos(3x)
Rewrite as=cos(2x+x)
Use the Angle Sum identity: cos(s+t)=cos(s)cos(t)−sin(s)sin(t)=cos(2x)cos(x)−sin(2x)sin(x)
Use the Double Angle identity: sin(2x)=2sin(x)cos(x)=cos(2x)cos(x)−2sin(x)cos(x)sin(x)
Simplify cos(2x)cos(x)−2sin(x)cos(x)sin(x):cos(x)cos(2x)−2sin2(x)cos(x)
cos(2x)cos(x)−2sin(x)cos(x)sin(x)
2sin(x)cos(x)sin(x)=2sin2(x)cos(x)
2sin(x)cos(x)sin(x)
Apply exponent rule: ab⋅ac=ab+csin(x)sin(x)=sin1+1(x)=2cos(x)sin1+1(x)
Add the numbers: 1+1=2=2cos(x)sin2(x)
=cos(x)cos(2x)−2sin2(x)cos(x)
=cos(x)cos(2x)−2sin2(x)cos(x)
=cos(x)cos(2x)−2sin2(x)cos(x)
Use the Double Angle identity: cos(2x)=2cos2(x)−1=(2cos2(x)−1)cos(x)−2sin2(x)cos(x)
Use the Pythagorean identity: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=(2cos2(x)−1)cos(x)−2(1−cos2(x))cos(x)
Expand (2cos2(x)−1)cos(x)−2(1−cos2(x))cos(x):4cos3(x)−3cos(x)
(2cos2(x)−1)cos(x)−2(1−cos2(x))cos(x)
=cos(x)(2cos2(x)−1)−2cos(x)(1−cos2(x))
Expand cos(x)(2cos2(x)−1):2cos3(x)−cos(x)
cos(x)(2cos2(x)−1)
Apply the distributive law: a(b−c)=ab−aca=cos(x),b=2cos2(x),c=1=cos(x)2cos2(x)−cos(x)1
=2cos2(x)cos(x)−1cos(x)
Simplify 2cos2(x)cos(x)−1⋅cos(x):2cos3(x)−cos(x)
2cos2(x)cos(x)−1cos(x)
2cos2(x)cos(x)=2cos3(x)
2cos2(x)cos(x)
Apply exponent rule: ab⋅ac=ab+ccos2(x)cos(x)=cos2+1(x)=2cos2+1(x)
Add the numbers: 2+1=3=2cos3(x)
1⋅cos(x)=cos(x)
1cos(x)
Multiply: 1⋅cos(x)=cos(x)=cos(x)
=2cos3(x)−cos(x)
=2cos3(x)−cos(x)
=2cos3(x)−cos(x)−2(1−cos2(x))cos(x)
Expand −2cos(x)(1−cos2(x)):−2cos(x)+2cos3(x)
−2cos(x)(1−cos2(x))
Apply the distributive law: a(b−c)=ab−aca=−2cos(x),b=1,c=cos2(x)=−2cos(x)1−(−2cos(x))cos2(x)
Apply minus-plus rules−(−a)=a=−2⋅1cos(x)+2cos2(x)cos(x)
Simplify −2⋅1⋅cos(x)+2cos2(x)cos(x):−2cos(x)+2cos3(x)
−2⋅1cos(x)+2cos2(x)cos(x)
2⋅1⋅cos(x)=2cos(x)
2⋅1cos(x)
Multiply the numbers: 2⋅1=2=2cos(x)
2cos2(x)cos(x)=2cos3(x)
2cos2(x)cos(x)
Apply exponent rule: ab⋅ac=ab+ccos2(x)cos(x)=cos2+1(x)=2cos2+1(x)
Add the numbers: 2+1=3=2cos3(x)
=−2cos(x)+2cos3(x)
=−2cos(x)+2cos3(x)
=2cos3(x)−cos(x)−2cos(x)+2cos3(x)
Simplify 2cos3(x)−cos(x)−2cos(x)+2cos3(x):4cos3(x)−3cos(x)
2cos3(x)−cos(x)−2cos(x)+2cos3(x)
Group like terms=2cos3(x)+2cos3(x)−cos(x)−2cos(x)
Add similar elements: 2cos3(x)+2cos3(x)=4cos3(x)=4cos3(x)−cos(x)−2cos(x)
Add similar elements: −cos(x)−2cos(x)=−3cos(x)=4cos3(x)−3cos(x)
=4cos3(x)−3cos(x)
=4cos3(x)−3cos(x)
=−1+4cos3(x)(4cos3(x)−3cos(x)+3cos(x))
4cos3(x)(4cos3(x)−3cos(x)+3cos(x))=16cos6(x)
4cos3(x)(4cos3(x)−3cos(x)+3cos(x))
Add similar elements: −3cos(x)+3cos(x)=0=4⋅4cos3(x)cos3(x)
Multiply the numbers: 4⋅4=16=16cos3(x)cos3(x)
Apply exponent rule: ab⋅ac=ab+ccos3(x)cos3(x)=cos3+3(x)=16cos3+3(x)
Add the numbers: 3+3=6=16cos6(x)
=−1+16cos6(x)
−1+16cos6(x)=0
Solve by substitution
−1+16cos6(x)=0
Let: cos(x)=w−1+16w6=0
−1+16w6=0
Move 1to the right side
−1+16w6=0
Add 1 to both sides−1+16w6+1=0+1
Simplify16w6=1
16w6=1
Divide both sides by 16
16w6=1
Divide both sides by 161616w6​=161​
Simplifyw6=161​
w6=161​
Rewrite the equation with u=w2 and u3=w6u3=161​
Solve
u3=161​
For g3(x)=f(a) the solutions are
Substitute back u=w2,solve for w
Solve
Simplify
Apply radical rule: assuming a≥0,b≥0
Prime factorization of 16:24
16
16divides by 216=8⋅2=2⋅8
8divides by 28=4⋅2=2⋅2⋅4
4divides by 24=2⋅2=2⋅2⋅2⋅2
2 is a prime number, therefore no further factorization is possible=2⋅2⋅2⋅2
=24
Apply exponent rule: ab+c=ab⋅ac
Apply radical rule:
Apply radical rule:
Apply rule
Rationalize
Multiply by the conjugate 232​232​​
1⋅232​=232​
Apply exponent rule: ab⋅ac=ab+c=21+32​+31​
Join 1+32​+31​:2
1+32​+31​
Convert element to fraction: 1=11​=11​+32​+31​
Least Common Multiplier of 1,3,3:3
1,3,3
Least Common Multiplier (LCM)
Prime factorization of 1
Prime factorization of 3:3
3
3 is a prime number, therefore no factorization is possible=3
Prime factorization of 3:3
3
3 is a prime number, therefore no factorization is possible=3
Compute a number comprised of factors that appear in at least one of the following:
1,3,3
=3
Multiply the numbers: 3=3=3
Adjust Fractions based on the LCM
Multiply each numerator by the same amount needed to multiply its
corresponding denominator to turn it into the LCM 3
For 11​:multiply the denominator and numerator by 311​=1⋅31⋅3​=33​
=33​+32​+31​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=33+2+1​
Add the numbers: 3+2+1=6=36​
Divide the numbers: 36​=2=2
=22
22=4=4
=4232​​
=4232​​
w2=4232​​
For x2=f(a) the solutions are x=f(a)​,−f(a)​
w=4232​​​,w=−4232​​​
4232​​​=2232​​​
4232​​​
Apply radical rule: assuming a≥0,b≥0=4​232​​​
4​=2
4​
Factor the number: 4=22=22​
Apply radical rule: 22​=2=2
=2232​​​
−4232​​​=−2232​​​
−4232​​​
Simplify 4232​​​:2232​​​
4232​​​
Apply radical rule: assuming a≥0,b≥0=4​232​​​
4​=2
4​
Factor the number: 4=22=22​
Apply radical rule: 22​=2=2
=2232​​​
=−2232​​​
w=2232​​​,w=−2232​​​
Solve
Substitute w=u+vi
Expand
Expand (u+vi)2:(u2−v2)+2iuv
(u+vi)2
=(u+iv)2
Apply Perfect Square Formula: (a+b)2=a2+2ab+b2a=u,b=vi
=u2+2uvi+(vi)2
(vi)2=−v2
(vi)2
Apply exponent rule: (a⋅b)n=anbn=i2v2
i2=−1
i2
Apply imaginary number rule: i2=−1=−1
=(−1)v2
Refine=−v2
=u2+2iuv−v2
Rewrite u2+2iuv−v2 in standard complex form: (u2−v2)+2uvi
u2+2iuv−v2
Group the real part and the imaginary part of the complex number=(u2−v2)+2uvi
=(u2−v2)+2uvi
Expand
Apply radical rule: assuming a≥0,b≥0
Prime factorization of 16:24
16
16divides by 216=8⋅2=2⋅8
8divides by 28=4⋅2=2⋅2⋅4
4divides by 24=2⋅2=2⋅2⋅2⋅2
2 is a prime number, therefore no further factorization is possible=2⋅2⋅2⋅2
=24
Apply exponent rule: ab+c=ab⋅ac
Apply radical rule:
Apply radical rule:
Apply rule
Multiply fractions: ba​⋅dc​=b⋅da⋅c​
1⋅(−1+3​i)=−1+3​i
1⋅(−1+3​i)
Multiply: 1⋅(−1+3​i)=(−1+3​i)=(−1+3​i)
Remove parentheses: (−a)=−a=−1+3​i
Multiply the numbers: 2⋅2=4
Apply the fraction rule: ca±b​=ca​±cb​
Rewrite in standard complex form: −8232​​+83​⋅232​​i
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​
Apply the fraction rule: ca±b​=ca​±cb​
Multiply by the conjugate 232​232​​
Apply exponent rule: ab⋅ac=ab+c=4⋅232​+31​
232​+31​=2
232​+31​
Combine the fractions 32​+31​:1
Apply rule ca​±cb​=ca±b​=32+1​
Add the numbers: 2+1=3=33​
Apply rule aa​=1=1
=21
Apply rule a1=a=2
=4⋅2
Multiply the numbers: 4⋅2=8=8
=83​⋅232​​
Multiply by the conjugate 232​232​​
1⋅232​=232​
Apply exponent rule: ab⋅ac=ab+c=4⋅232​+31​
232​+31​=2
232​+31​
Combine the fractions 32​+31​:1
Apply rule ca​±cb​=ca±b​=32+1​
Add the numbers: 2+1=3=33​
Apply rule aa​=1=1
=21
Apply rule a1=a=2
=4⋅2
Multiply the numbers: 4⋅2=8=8
=−8232​​
=−8232​​+83​⋅232​​i
=−8232​​+83​⋅232​​i
(u2−v2)+2iuv=−8232​​+i8232​3​​
(u2−v2)+2iuv=−8232​​+i8232​3​​
Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:[u2−v2=−8232​​2uv=83​⋅232​​​]
[u2−v2=−8232​​2uv=83​⋅232​​​]
Isolate ufor 2uv=8232​⋅3​​:u=2310​v3​​
2uv=8232​3​​
Simplify 8232​3​​:237​3​​
8232​3​​
Factor the number: 8=23=23232​3​​
Simplify 23232​​:237​1​
23232​​
Apply exponent rule: xbxa​=xb−a1​=23−32​1​
3−32​=37​
3−32​
Convert element to fraction: 3=33⋅3​=33⋅3​−32​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=33⋅3−2​
3⋅3−2=7
3⋅3−2
Multiply the numbers: 3⋅3=9=9−2
Subtract the numbers: 9−2=7=7
=37​
=237​1​
=237​3​​
2uv=237​3​​
Divide both sides by 2v
2uv=237​3​​
Divide both sides by 2v2v2uv​=2v237​3​​​
Simplify
2v2uv​=2v237​3​​​
Simplify 2v2uv​:u
2v2uv​
Cancel the common factor: 2=vuv​
Cancel the common factor: v=u
Simplify 2v237​3​​​:2310​v3​​
2v237​3​​​
Apply the fraction rule: cba​​=b⋅ca​=237​⋅2v3​​
237​⋅2=2310​
237​⋅2
Apply exponent rule: ab⋅ac=ab+c237​⋅2=237​+1=237​+1
37​+1=310​
37​+1
Convert element to fraction: 1=31⋅3​=37​+31⋅3​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=37+1⋅3​
7+1⋅3=10
7+1⋅3
Multiply the numbers: 1⋅3=3=7+3
Add the numbers: 7+3=10=10
=310​
=2310​
=2310​v3​​
u=2310​v3​​
u=2310​v3​​
u=2310​v3​​
Plug the solutions u=2310​v3​​ into u2−v2=−8232​​
For u2−v2=−8232​​, subsitute u with 2310​v3​​:v≈0.54556…,v≈−0.54556…
For u2−v2=−8232​​, subsitute u with 2310​v3​​(2310​v3​​)2−v2=−8232​​
Solve (2310​v3​​)2−v2=−8232​​:v≈0.54556…,v≈−0.54556…
(2310​v3​​)2−v2=−8232​​
Multiply by LCM
(2310​v3​​)2−v2=−8232​​
Simplify (2310​v3​​)2:26⋅232​v23​
(2310​v3​​)2
2310​v3​​
2310​
2310​=23+31​=23+31​
Apply exponent rule: xa+b=xaxb=23⋅231​
Refine
Apply exponent rule: (ba​)c=bcac​
Apply exponent rule: (a⋅b)n=anbn
(3​)2:3
Apply radical rule: a​=a21​=(321​)2
Apply exponent rule: (ab)c=abc=321​⋅2
21​⋅2=1
21​⋅2
Multiply fractions: a⋅cb​=ca⋅b​=21⋅2​
Cancel the common factor: 2=1
=3
(23)2:26
Apply exponent rule: (ab)c=abc=23⋅2
Multiply the numbers: 3⋅2=6=26
Apply radical rule: =(231​)2
Apply exponent rule: (ab)c=abc=231​⋅2
31​⋅2=32​
31​⋅2
Multiply fractions: a⋅cb​=ca⋅b​=31⋅2​
Multiply the numbers: 1⋅2=2=32​
=232​
=26⋅232​v23​
26⋅232​v2=26+32​v2
26⋅232​v2
Apply exponent rule: ab⋅ac=ab+c26⋅232​=26+32​=26+32​v2
=232​+6v23​
26+32​=26⋅232​
26+32​
Apply exponent rule: xa+b=xaxb=26⋅232​
=26⋅232​v23​
26⋅232​v23​−v2=−8232​​
Find Least Common Multiplier of 26+32​v2,8:64⋅232​v2
26+32​v2,8
Lowest Common Multiplier (LCM)
Least Common Multiplier of 64,8:64
64,8
Least Common Multiplier (LCM)
Prime factorization of 64:2⋅2⋅2⋅2⋅2⋅2
64
64divides by 264=32⋅2=2⋅32
32divides by 232=16⋅2=2⋅2⋅16
16divides by 216=8⋅2=2⋅2⋅2⋅8
8divides by 28=4⋅2=2⋅2⋅2⋅2⋅4
4divides by 24=2⋅2=2⋅2⋅2⋅2⋅2⋅2
Prime factorization of 8:2⋅2⋅2
8
8divides by 28=4⋅2=2⋅4
4divides by 24=2⋅2=2⋅2⋅2
Multiply each factor the greatest number of times it occurs in either 64 or 8=2⋅2⋅2⋅2⋅2⋅2
Multiply the numbers: 2⋅2⋅2⋅2⋅2⋅2=64=64
Compute an expression comprised of factors that appear either in 64⋅232​v2 or 8=64⋅232​v2
Multiply by LCM=64⋅232​v226⋅232​v23​⋅64⋅232​v2−v2⋅64⋅232​v2=−8232​​⋅64⋅232​v2
Simplify
26⋅232​v23​⋅64⋅232​v2−v2⋅64⋅232​v2=−8232​​⋅64⋅232​v2
Simplify 26⋅232​v23​⋅64⋅232​v2:3
26⋅232​v23​⋅64⋅232​v2
Multiply fractions: a⋅cb​=ca⋅b​=26⋅232​v23⋅64⋅232​v2​
Cancel the common factor: 232​=26v23⋅64v2​
Cancel the common factor: v2=263⋅64​
Multiply the numbers: 3⋅64=192=26192​
Factor 192:26⋅3
Factor 192=26⋅3
=2626⋅3​
Cancel the common factor: 26=3
Simplify −v2⋅64⋅232​v2:−64⋅232​v4
−v2⋅64⋅232​v2
Apply exponent rule: ab⋅ac=ab+cv2v2=v2+2=−64⋅232​v2+2
Add the numbers: 2+2=4=−64⋅232​v4
Simplify
−8232​​⋅64⋅232​v2
Multiply fractions: a⋅cb​=ca⋅b​=−8232​⋅64⋅232​v2​
232​⋅64⋅232​v2=64⋅22⋅32​v2
232​⋅64⋅232​v2
Apply exponent rule: ab⋅ac=ab+c232​⋅232​=232​+32​=64⋅232​+32​v2
Add similar elements: 32​+32​=2⋅32​=64⋅22⋅32​v2
=−864⋅22⋅32​v2​
Divide the numbers: 864​=8=−8⋅22⋅32​v2
22⋅32​
Multiply 2⋅32​:34​
2⋅32​
Multiply fractions: a⋅cb​=ca⋅b​=32⋅2​
Multiply the numbers: 2⋅2=4=34​
=234​
234​=21+31​=21+31​
Apply exponent rule: xa+b=xaxb=21⋅231​
Refine
Multiply the numbers: 8⋅2=16
Solve
Move to the left side
Add to both sides
Simplify
Write in the standard form an​xn+…+a1​x+a0​=0
Find one solution for −101.59366…v4+20.15873…v2+3=0 using Newton-Raphson:v≈0.54556…
−101.59366…v4+20.15873…v2+3=0
Newton-Raphson Approximation Definition
f(v)=−101.59366…v4+20.15873…v2+3
Find f′(v):−406.37466…v3+40.31747…v
dvd​(−101.59366…v4+20.15873…v2+3)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=−dvd​(101.59366…v4)+dvd​(20.15873…v2)+dvd​(3)
dvd​(101.59366…v4)=406.37466…v3
dvd​(101.59366…v4)
Take the constant out: (a⋅f)′=a⋅f′=101.59366…dvd​(v4)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=101.59366…⋅4v4−1
Simplify=406.37466…v3
dvd​(20.15873…v2)=40.31747…v
dvd​(20.15873…v2)
Take the constant out: (a⋅f)′=a⋅f′=20.15873…dvd​(v2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=20.15873…⋅2v2−1
Simplify=40.31747…v
dvd​(3)=0
dvd​(3)
Derivative of a constant: dxd​(a)=0=0
=−406.37466…v3+40.31747…v+0
Simplify=−406.37466…v3+40.31747…v
Let v0​=1Compute vn+1​ until Δvn+1​<0.000001
v1​=0.78573…:Δv1​=0.21426…
f(v0​)=−101.59366…⋅14+20.15873…⋅12+3=−78.43493…f′(v0​)=−406.37466…⋅13+40.31747…⋅1=−366.05719…v1​=0.78573…
Δv1​=∣0.78573…−1∣=0.21426…Δv1​=0.21426…
v2​=0.64504…:Δv2​=0.14068…
f(v1​)=−101.59366…⋅0.78573…4+20.15873…⋅0.78573…2+3=−23.27682…f′(v1​)=−406.37466…⋅0.78573…3+40.31747…⋅0.78573…=−165.44887…v2​=0.64504…
Δv2​=∣0.64504…−0.78573…∣=0.14068…Δv2​=0.14068…
v3​=0.57039…:Δv3​=0.07465…
f(v2​)=−101.59366…⋅0.64504…4+20.15873…⋅0.64504…2+3=−6.20040…f′(v2​)=−406.37466…⋅0.64504…3+40.31747…⋅0.64504…=−83.05958…v3​=0.57039…
Δv3​=∣0.57039…−0.64504…∣=0.07465…Δv3​=0.07465…
v4​=0.54759…:Δv4​=0.02280…
f(v3​)=−101.59366…⋅0.57039…4+20.15873…⋅0.57039…2+3=−1.19513…f′(v3​)=−406.37466…⋅0.57039…3+40.31747…⋅0.57039…=−52.41610…v4​=0.54759…
Δv4​=∣0.54759…−0.57039…∣=0.02280…Δv4​=0.02280…
v5​=0.54557…:Δv5​=0.00201…
f(v4​)=−101.59366…⋅0.54759…4+20.15873…⋅0.54759…2+3=−0.08990…f′(v4​)=−406.37466…⋅0.54759…3+40.31747…⋅0.54759…=−44.64836…v5​=0.54557…
Δv5​=∣0.54557…−0.54759…∣=0.00201…Δv5​=0.00201…
v6​=0.54556…:Δv6​=0.00001…
f(v5​)=−101.59366…⋅0.54557…4+20.15873…⋅0.54557…2+3=−0.00065…f′(v5​)=−406.37466…⋅0.54557…3+40.31747…⋅0.54557…=−43.99616…v6​=0.54556…
Δv6​=∣0.54556…−0.54557…∣=0.00001…Δv6​=0.00001…
v7​=0.54556…:Δv7​=8.18838E−10
f(v6​)=−101.59366…⋅0.54556…4+20.15873…⋅0.54556…2+3=−3.60218E−8f′(v6​)=−406.37466…⋅0.54556…3+40.31747…⋅0.54556…=−43.99134…v7​=0.54556…
Δv7​=∣0.54556…−0.54556…∣=8.18838E−10Δv7​=8.18838E−10
v≈0.54556…
Apply long division:
−101.59366…v3−55.42562…v2−10.07936…v−5.49891…≈0
Find one solution for −101.59366…v3−55.42562…v2−10.07936…v−5.49891…=0 using Newton-Raphson:v≈−0.54556…
−101.59366…v3−55.42562…v2−10.07936…v−5.49891…=0
Newton-Raphson Approximation Definition
f(v)=−101.59366…v3−55.42562…v2−10.07936…v−5.49891…
Find f′(v):−304.78100…v2−110.85125…v−10.07936…
dvd​(−101.59366…v3−55.42562…v2−10.07936…v−5.49891…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=−dvd​(101.59366…v3)−dvd​(55.42562…v2)−dvd​(10.07936…v)−dvd​(5.49891…)
dvd​(101.59366…v3)=304.78100…v2
dvd​(101.59366…v3)
Take the constant out: (a⋅f)′=a⋅f′=101.59366…dvd​(v3)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=101.59366…⋅3v3−1
Simplify=304.78100…v2
dvd​(55.42562…v2)=110.85125…v
dvd​(55.42562…v2)
Take the constant out: (a⋅f)′=a⋅f′=55.42562…dvd​(v2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=55.42562…⋅2v2−1
Simplify=110.85125…v
dvd​(10.07936…v)=10.07936…
dvd​(10.07936…v)
Take the constant out: (a⋅f)′=a⋅f′=10.07936…dvdv​
Apply the common derivative: dvdv​=1=10.07936…⋅1
Simplify=10.07936…
dvd​(5.49891…)=0
dvd​(5.49891…)
Derivative of a constant: dxd​(a)=0=0
=−304.78100…v2−110.85125…v−10.07936…−0
Simplify=−304.78100…v2−110.85125…v−10.07936…
Let v0​=−1Compute vn+1​ until Δvn+1​<0.000001
v1​=−0.75124…:Δv1​=0.24875…
f(v0​)=−101.59366…(−1)3−55.42562…(−1)2−10.07936…(−1)−5.49891…=50.74849…f′(v0​)=−304.78100…(−1)2−110.85125…(−1)−10.07936…=−204.00911…v1​=−0.75124…
Δv1​=∣−0.75124…−(−1)∣=0.24875…Δv1​=0.24875…
v2​=−0.61091…:Δv2​=0.14032…
f(v1​)=−101.59366…(−0.75124…)3−55.42562…(−0.75124…)2−10.07936…(−0.75124…)−5.49891…=13.86617…f′(v1​)=−304.78100…(−0.75124…)2−110.85125…(−0.75124…)−10.07936…=−98.81153…v2​=−0.61091…
Δv2​=∣−0.61091…−(−0.75124…)∣=0.14032…Δv2​=0.14032…
v3​=−0.55501…:Δv3​=0.05590…
f(v2​)=−101.59366…(−0.61091…)3−55.42562…(−0.61091…)2−10.07936…(−0.61091…)−5.49891…=3.13665…f′(v2​)=−304.78100…(−0.61091…)2−110.85125…(−0.61091…)−10.07936…=−56.10803…v3​=−0.55501…
Δv3​=∣−0.55501…−(−0.61091…)∣=0.05590…Δv3​=0.05590…
v4​=−0.54579…:Δv4​=0.00921…
f(v3​)=−101.59366…(−0.55501…)3−55.42562…(−0.55501…)2−10.07936…(−0.55501…)−5.49891…=0.39093…f′(v3​)=−304.78100…(−0.55501…)2−110.85125…(−0.55501…)−10.07936…=−42.43951…v4​=−0.54579…
Δv4​=∣−0.54579…−(−0.55501…)∣=0.00921…Δv4​=0.00921…
v5​=−0.54556…:Δv5​=0.00023…
f(v4​)=−101.59366…(−0.54579…)3−55.42562…(−0.54579…)2−10.07936…(−0.54579…)−5.49891…=0.00957…f′(v4​)=−304.78100…(−0.54579…)2−110.85125…(−0.54579…)−10.07936…=−40.37008…v5​=−0.54556…
Δv5​=∣−0.54556…−(−0.54579…)∣=0.00023…Δv5​=0.00023…
v6​=−0.54556…:Δv6​=1.54611E−7
f(v5​)=−101.59366…(−0.54556…)3−55.42562…(−0.54556…)2−10.07936…(−0.54556…)−5.49891…=6.23352E−6f′(v5​)=−304.78100…(−0.54556…)2−110.85125…(−0.54556…)−10.07936…=−40.31750…v6​=−0.54556…
Δv6​=∣−0.54556…−(−0.54556…)∣=1.54611E−7Δv6​=1.54611E−7
v≈−0.54556…
Apply long division:v+0.54556…−101.59366…v3−55.42562…v2−10.07936…v−5.49891…​=−101.59366…v2−10.07936…
−101.59366…v2−10.07936…≈0
Find one solution for −101.59366…v2−10.07936…=0 using Newton-Raphson:No Solution for v∈R
−101.59366…v2−10.07936…=0
Newton-Raphson Approximation Definition
f(v)=−101.59366…v2−10.07936…
Find f′(v):−203.18733…v
dvd​(−101.59366…v2−10.07936…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=−dvd​(101.59366…v2)−dvd​(10.07936…)
dvd​(101.59366…v2)=203.18733…v
dvd​(101.59366…v2)
Take the constant out: (a⋅f)′=a⋅f′=101.59366…dvd​(v2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=101.59366…⋅2v2−1
Simplify=203.18733…v
dvd​(10.07936…)=0
dvd​(10.07936…)
Derivative of a constant: dxd​(a)=0=0
=−203.18733…v−0
Simplify=−203.18733…v
Let v0​=−1Compute vn+1​ until Δvn+1​<0.000001
v1​=−0.45039…:Δv1​=0.54960…
f(v0​)=−101.59366…(−1)2−10.07936…=−111.67303…f′(v0​)=−203.18733…(−1)=203.18733…v1​=−0.45039…
Δv1​=∣−0.45039…−(−1)∣=0.54960…Δv1​=0.54960…
v2​=−0.11505…:Δv2​=0.33533…
f(v1​)=−101.59366…(−0.45039…)2−10.07936…=−30.68810…f′(v1​)=−203.18733…(−0.45039…)=91.51429…v2​=−0.11505…
Δv2​=∣−0.11505…−(−0.45039…)∣=0.33533…Δv2​=0.33533…
v3​=0.37361…:Δv3​=0.48867…
f(v2​)=−101.59366…(−0.11505…)2−10.07936…=−11.42427…f′(v2​)=−203.18733…(−0.11505…)=23.37813…v3​=0.37361…
Δv3​=∣0.37361…−(−0.11505…)∣=0.48867…Δv3​=0.48867…
v4​=0.05403…:Δv4​=0.31958…
f(v3​)=−101.59366…⋅0.37361…2−10.07936…=−24.26076…f′(v3​)=−203.18733…⋅0.37361…=−75.91416…v4​=0.05403…
Δv4​=∣0.05403…−0.37361…∣=0.31958…Δv4​=0.31958…
v5​=−0.89102…:Δv5​=0.94505…
f(v4​)=−101.59366…⋅0.05403…2−10.07936…=−10.37600…f′(v4​)=−203.18733…⋅0.05403…=−10.97924…v5​=−0.89102…
Δv5​=∣−0.89102…−0.05403…∣=0.94505…Δv5​=0.94505…
v6​=−0.38983…:Δv6​=0.50118…
f(v5​)=−101.59366…(−0.89102…)2−10.07936…=−90.73642…f′(v5​)=−203.18733…(−0.89102…)=181.04414…v6​=−0.38983…
Δv6​=∣−0.38983…−(−0.89102…)∣=0.50118…Δv6​=0.50118…
v7​=−0.06766…:Δv7​=0.32216…
f(v6​)=−101.59366…(−0.38983…)2−10.07936…=−25.51884…f′(v6​)=−203.18733…(−0.38983…)=79.20991…v7​=−0.06766…
Δv7​=∣−0.06766…−(−0.38983…)∣=0.32216…Δv7​=0.32216…
v8​=0.69923…:Δv8​=0.76690…
f(v7​)=−101.59366…(−0.06766…)2−10.07936…=−10.54458…f′(v7​)=−203.18733…(−0.06766…)=13.74960…v8​=0.69923…
Δv8​=∣0.69923…−(−0.06766…)∣=0.76690…Δv8​=0.76690…
v9​=0.27867…:Δv9​=0.42055…
f(v8​)=−101.59366…⋅0.69923…2−10.07936…=−59.75094…f′(v8​)=−203.18733…⋅0.69923…=−142.07487…v9​=0.27867…
Δv9​=∣0.27867…−0.69923…∣=0.42055…Δv9​=0.42055…
v10​=−0.03867…:Δv10​=0.31734…
f(v9​)=−101.59366…⋅0.27867…2−10.07936…=−17.96890…f′(v9​)=−203.18733…⋅0.27867…=−56.62250…v10​=−0.03867…
Δv10​=∣−0.03867…−0.27867…∣=0.31734…Δv10​=0.31734…
v11​=1.26333…:Δv11​=1.30200…
f(v10​)=−101.59366…(−0.03867…)2−10.07936…=−10.23132…f′(v10​)=−203.18733…(−0.03867…)=7.85811…v11​=1.26333…
Δv11​=∣1.26333…−(−0.03867…)∣=1.30200…Δv11​=1.30200…
Cannot find solution
The solutions arev≈0.54556…,v≈−0.54556…
v≈0.54556…,v≈−0.54556…
Verify Solutions
Find undefined (singularity) points:v=0
Take the denominator(s) of (2310​v3​​)2−v2 and compare to zero
Solve 2310​v=0:v=0
2310​v=0
Divide both sides by 2310​
2310​v=0
Divide both sides by 2310​2310​2310​v​=2310​0​
Simplify
2310​2310​v​=2310​0​
Simplify 2310​2310​v​:v
2310​2310​v​
Cancel the common factor: 2310​=v
Simplify 2310​0​:0
2310​0​
Apply rule a0​=0: a=0=0
v=0
v=0
v=0
The following points are undefinedv=0
Combine undefined points with solutions:
v≈0.54556…,v≈−0.54556…
Plug the solutions v=0.54556…,v=−0.54556… into 2uv=8232​⋅3​​
For 2uv=8232​⋅3​​, subsitute v with
For 2uv=8232​⋅3​​, subsitute v with 0.54556…2u⋅0.54556…=8232​3​​
Solve
2u⋅0.54556…=8232​3​​
Simplify 8232​3​​:237​3​​
8232​3​​
Factor the number: 8=23=23232​3​​
Simplify 23232​​:237​1​
23232​​
Apply exponent rule: xbxa​=xb−a1​=23−32​1​
3−32​=37​
3−32​
Convert element to fraction: 3=33⋅3​=33⋅3​−32​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=33⋅3−2​
3⋅3−2=7
3⋅3−2
Multiply the numbers: 3⋅3=9=9−2
Subtract the numbers: 9−2=7=7
=37​
=237​1​
=237​3​​
2u⋅0.54556…=237​3​​
Divide both sides by 2⋅0.54556…
2u⋅0.54556…=237​3​​
Divide both sides by 2⋅0.54556…2⋅0.54556…2u⋅0.54556…​=2⋅0.54556…237​3​​​
Simplify
2⋅0.54556…2u⋅0.54556…​=2⋅0.54556…237​3​​​
Simplify 2⋅0.54556…2u⋅0.54556…​:u
2⋅0.54556…2u⋅0.54556…​
Cancel the common factor: 2=0.54556…u⋅0.54556…​
Cancel the common factor: 0.54556…=u
Simplify
2⋅0.54556…237​3​​​
Multiply the numbers: 2⋅0.54556…=1.09112…=1.09112…237​3​​​
Apply the fraction rule: acb​​=c⋅ab​=237​⋅1.09112…3​​
237​
237​=22+31​=22+31​
Apply exponent rule: xa+b=xaxb=22⋅231​
Refine
For 2uv=8232​⋅3​​, subsitute v with
For 2uv=8232​⋅3​​, subsitute v with −0.54556…2u(−0.54556…)=8232​3​​
Solve
2u(−0.54556…)=8232​3​​
Simplify 8232​3​​:237​3​​
8232​3​​
Factor the number: 8=23=23232​3​​
Simplify 23232​​:237​1​
23232​​
Apply exponent rule: xbxa​=xb−a1​=23−32​1​
3−32​=37​
3−32​
Convert element to fraction: 3=33⋅3​=33⋅3​−32​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=33⋅3−2​
3⋅3−2=7
3⋅3−2
Multiply the numbers: 3⋅3=9=9−2
Subtract the numbers: 9−2=7=7
=37​
=237​1​
=237​3​​
2u(−0.54556…)=237​3​​
Divide both sides by 2(−0.54556…)
2u(−0.54556…)=237​3​​
Divide both sides by 2(−0.54556…)2(−0.54556…)2u(−0.54556…)​=2(−0.54556…)237​3​​​
Simplify
2(−0.54556…)2u(−0.54556…)​=2(−0.54556…)237​3​​​
Simplify 2(−0.54556…)2u(−0.54556…)​:u
2(−0.54556…)2u(−0.54556…)​
Multiply the numbers: 2(−0.54556…)=−1.09112…=−1.09112…−1.09112…u​
Cancel the common factor: −1.09112…=u
Simplify
2(−0.54556…)237​3​​​
Remove parentheses: (−a)=−a=−2⋅0.54556…237​3​​​
Multiply the numbers: 2⋅0.54556…=1.09112…=−1.09112…237​3​​​
Apply the fraction rule: −ba​=−ba​=−1.09112…237​3​​​
Apply the fraction rule: acb​​=c⋅ab​1.09112…237​3​​​=237​⋅1.09112…3​​=−237​⋅1.09112…3​​
237​
237​=22+31​=22+31​
Apply exponent rule: xa+b=xaxb=22⋅231​
Refine
Verify solutions by plugging them into the original equations
Check the solutions by plugging them into u2−v2=−8232​​
Remove the ones that don't agree with the equation.
Check the solution True
u2−v2=−8232​​
Plug in
Refine
True
Check the solution True
u2−v2=−8232​​
Plug in
Refine
True
Check the solutions by plugging them into 2uv=83​⋅232​​
Remove the ones that don't agree with the equation.
Check the solution True
2uv=83​⋅232​​
Plug in
Refine4.36449…3​⋅232​⋅0.54556…​=83​⋅232​​
True
Check the solution True
2uv=83​⋅232​​
Plug in
Refine4.36449…3​⋅232​⋅0.54556…​=83​⋅232​​
True
Therefore, the final solutions for u2−v2=−8232​​,2uv=83​⋅232​​ are
Substitute back w=u+vi
Solve
Substitute w=u+vi
Expand
Expand (u+vi)2:(u2−v2)+2iuv
(u+vi)2
=(u+iv)2
Apply Perfect Square Formula: (a+b)2=a2+2ab+b2a=u,b=vi
=u2+2uvi+(vi)2
(vi)2=−v2
(vi)2
Apply exponent rule: (a⋅b)n=anbn=i2v2
i2=−1
i2
Apply imaginary number rule: i2=−1=−1
=(−1)v2
Refine=−v2
=u2+2iuv−v2
Rewrite u2+2iuv−v2 in standard complex form: (u2−v2)+2uvi
u2+2iuv−v2
Group the real part and the imaginary part of the complex number=(u2−v2)+2uvi
=(u2−v2)+2uvi
Expand
Apply radical rule: assuming a≥0,b≥0
Prime factorization of 16:24
16
16divides by 216=8⋅2=2⋅8
8divides by 28=4⋅2=2⋅2⋅4
4divides by 24=2⋅2=2⋅2⋅2⋅2
2 is a prime number, therefore no further factorization is possible=2⋅2⋅2⋅2
=24
Apply exponent rule: ab+c=ab⋅ac
Apply radical rule:
Apply radical rule:
Apply rule
Multiply fractions: ba​⋅dc​=b⋅da⋅c​
1⋅(−1−3​i)=−1−3​i
1⋅(−1−3​i)
Multiply: 1⋅(−1−3​i)=(−1−3​i)=(−1−3​i)
Remove parentheses: (−a)=−a=−1−3​i
Multiply the numbers: 2⋅2=4
Apply the fraction rule: ca±b​=ca​±cb​
Rewrite in standard complex form: −8232​​−83​⋅232​​i
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​
Apply the fraction rule: ca±b​=ca​±cb​
Multiply by the conjugate 232​232​​
Apply exponent rule: ab⋅ac=ab+c=4⋅232​+31​
232​+31​=2
232​+31​
Combine the fractions 32​+31​:1
Apply rule ca​±cb​=ca±b​=32+1​
Add the numbers: 2+1=3=33​
Apply rule aa​=1=1
=21
Apply rule a1=a=2
=4⋅2
Multiply the numbers: 4⋅2=8=8
=−83​⋅232​​
Multiply by the conjugate 232​232​​
1⋅232​=232​
Apply exponent rule: ab⋅ac=ab+c=4⋅232​+31​
232​+31​=2
232​+31​
Combine the fractions 32​+31​:1
Apply rule ca​±cb​=ca±b​=32+1​
Add the numbers: 2+1=3=33​
Apply rule aa​=1=1
=21
Apply rule a1=a=2
=4⋅2
Multiply the numbers: 4⋅2=8=8
=−8232​​
=−8232​​−83​⋅232​​i
=−8232​​−83​⋅232​​i
(u2−v2)+2iuv=−8232​​−i8232​3​​
(u2−v2)+2iuv=−8232​​−i8232​3​​
Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:[u2−v2=−8232​​2uv=−83​⋅232​​​]
[u2−v2=−8232​​2uv=−83​⋅232​​​]
Isolate ufor 2uv=−8232​⋅3​​:u=−2310​v3​​
2uv=−8232​3​​
Simplify −8232​3​​:−237​3​​
−8232​3​​
Factor the number: 8=23=−23232​3​​
Simplify 23232​​:237​1​
23232​​
Apply exponent rule: xbxa​=xb−a1​=23−32​1​
3−32​=37​
3−32​
Convert element to fraction: 3=33⋅3​=33⋅3​−32​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=33⋅3−2​
3⋅3−2=7
3⋅3−2
Multiply the numbers: 3⋅3=9=9−2
Subtract the numbers: 9−2=7=7
=37​
=237​1​
=−237​3​​
2uv=−237​3​​
Divide both sides by 2v
2uv=−237​3​​
Divide both sides by 2v2v2uv​=2v−237​3​​​
Simplify
2v2uv​=2v−237​3​​​
Simplify 2v2uv​:u
2v2uv​
Cancel the common factor: 2=vuv​
Cancel the common factor: v=u
Simplify 2v−237​3​​​:−2310​v3​​
2v−237​3​​​
Apply the fraction rule: b−a​=−ba​=−2v237​3​​​
Apply the fraction rule: cba​​=b⋅ca​2v237​3​​​=237​⋅2v3​​=−237​⋅2v3​​
237​⋅2=2310​
237​⋅2
Apply exponent rule: ab⋅ac=ab+c237​⋅2=237​+1=237​+1
37​+1=310​
37​+1
Convert element to fraction: 1=31⋅3​=37​+31⋅3​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=37+1⋅3​
7+1⋅3=10
7+1⋅3
Multiply the numbers: 1⋅3=3=7+3
Add the numbers: 7+3=10=10
=310​
=2310​
=−2310​v3​​
u=−2310​v3​​
u=−2310​v3​​
u=−2310​v3​​
Plug the solutions u=−2310​v3​​ into u2−v2=−8232​​
For u2−v2=−8232​​, subsitute u with −2310​v3​​:v≈0.54556…,v≈−0.54556…
For u2−v2=−8232​​, subsitute u with −2310​v3​​(−2310​v3​​)2−v2=−8232​​
Solve (−2310​v3​​)2−v2=−8232​​:v≈0.54556…,v≈−0.54556…
(−2310​v3​​)2−v2=−8232​​
Multiply by LCM
(−2310​v3​​)2−v2=−8232​​
Simplify (−2310​v3​​)2:26⋅232​v23​
(−2310​v3​​)2
2310​v3​​
2310​
2310​=23+31​=23+31​
Apply exponent rule: xa+b=xaxb=23⋅231​
Refine
Apply exponent rule: (−a)n=an,if n is even
Apply exponent rule: (ba​)c=bcac​
Apply exponent rule: (a⋅b)n=anbn
(3​)2:3
Apply radical rule: a​=a21​=(321​)2
Apply exponent rule: (ab)c=abc=321​⋅2
21​⋅2=1
21​⋅2
Multiply fractions: a⋅cb​=ca⋅b​=21⋅2​
Cancel the common factor: 2=1
=3
(23)2:26
Apply exponent rule: (ab)c=abc=23⋅2
Multiply the numbers: 3⋅2=6=26
Apply radical rule: =(231​)2
Apply exponent rule: (ab)c=abc=231​⋅2
31​⋅2=32​
31​⋅2
Multiply fractions: a⋅cb​=ca⋅b​=31⋅2​
Multiply the numbers: 1⋅2=2=32​
=232​
=26⋅232​v23​
26⋅232​v2=26+32​v2
26⋅232​v2
Apply exponent rule: ab⋅ac=ab+c26⋅232​=26+32​=26+32​v2
=232​+6v23​
26+32​=26⋅232​
26+32​
Apply exponent rule: xa+b=xaxb=26⋅232​
=26⋅232​v23​
26⋅232​v23​−v2=−8232​​
Find Least Common Multiplier of 26+32​v2,8:64⋅232​v2
26+32​v2,8
Lowest Common Multiplier (LCM)
Least Common Multiplier of 64,8:64
64,8
Least Common Multiplier (LCM)
Prime factorization of 64:2⋅2⋅2⋅2⋅2⋅2
64
64divides by 264=32⋅2=2⋅32
32divides by 232=16⋅2=2⋅2⋅16
16divides by 216=8⋅2=2⋅2⋅2⋅8
8divides by 28=4⋅2=2⋅2⋅2⋅2⋅4
4divides by 24=2⋅2=2⋅2⋅2⋅2⋅2⋅2
Prime factorization of 8:2⋅2⋅2
8
8divides by 28=4⋅2=2⋅4
4divides by 24=2⋅2=2⋅2⋅2
Multiply each factor the greatest number of times it occurs in either 64 or 8=2⋅2⋅2⋅2⋅2⋅2
Multiply the numbers: 2⋅2⋅2⋅2⋅2⋅2=64=64
Compute an expression comprised of factors that appear either in 64⋅232​v2 or 8=64⋅232​v2
Multiply by LCM=64⋅232​v226⋅232​v23​⋅64⋅232​v2−v2⋅64⋅232​v2=−8232​​⋅64⋅232​v2
Simplify
26⋅232​v23​⋅64⋅232​v2−v2⋅64⋅232​v2=−8232​​⋅64⋅232​v2
Simplify 26⋅232​v23​⋅64⋅232​v2:3
26⋅232​v23​⋅64⋅232​v2
Multiply fractions: a⋅cb​=ca⋅b​=26⋅232​v23⋅64⋅232​v2​
Cancel the common factor: 232​=26v23⋅64v2​
Cancel the common factor: v2=263⋅64​
Multiply the numbers: 3⋅64=192=26192​
Factor 192:26⋅3
Factor 192=26⋅3
=2626⋅3​
Cancel the common factor: 26=3
Simplify −v2⋅64⋅232​v2:−64⋅232​v4
−v2⋅64⋅232​v2
Apply exponent rule: ab⋅ac=ab+cv2v2=v2+2=−64⋅232​v2+2
Add the numbers: 2+2=4=−64⋅232​v4
Simplify
−8232​​⋅64⋅232​v2
Multiply fractions: a⋅cb​=ca⋅b​=−8232​⋅64⋅232​v2​
232​⋅64⋅232​v2=64⋅22⋅32​v2
232​⋅64⋅232​v2
Apply exponent rule: ab⋅ac=ab+c232​⋅232​=232​+32​=64⋅232​+32​v2
Add similar elements: 32​+32​=2⋅32​=64⋅22⋅32​v2
=−864⋅22⋅32​v2​
Divide the numbers: 864​=8=−8⋅22⋅32​v2
22⋅32​
Multiply 2⋅32​:34​
2⋅32​
Multiply fractions: a⋅cb​=ca⋅b​=32⋅2​
Multiply the numbers: 2⋅2=4=34​
=234​
234​=21+31​=21+31​
Apply exponent rule: xa+b=xaxb=21⋅231​
Refine
Multiply the numbers: 8⋅2=16
Solve
Move to the left side
Add to both sides
Simplify
Write in the standard form an​xn+…+a1​x+a0​=0
Find one solution for −101.59366…v4+20.15873…v2+3=0 using Newton-Raphson:v≈0.54556…
−101.59366…v4+20.15873…v2+3=0
Newton-Raphson Approximation Definition
f(v)=−101.59366…v4+20.15873…v2+3
Find f′(v):−406.37466…v3+40.31747…v
dvd​(−101.59366…v4+20.15873…v2+3)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=−dvd​(101.59366…v4)+dvd​(20.15873…v2)+dvd​(3)
dvd​(101.59366…v4)=406.37466…v3
dvd​(101.59366…v4)
Take the constant out: (a⋅f)′=a⋅f′=101.59366…dvd​(v4)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=101.59366…⋅4v4−1
Simplify=406.37466…v3
dvd​(20.15873…v2)=40.31747…v
dvd​(20.15873…v2)
Take the constant out: (a⋅f)′=a⋅f′=20.15873…dvd​(v2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=20.15873…⋅2v2−1
Simplify=40.31747…v
dvd​(3)=0
dvd​(3)
Derivative of a constant: dxd​(a)=0=0
=−406.37466…v3+40.31747…v+0
Simplify=−406.37466…v3+40.31747…v
Let v0​=1Compute vn+1​ until Δvn+1​<0.000001
v1​=0.78573…:Δv1​=0.21426…
f(v0​)=−101.59366…⋅14+20.15873…⋅12+3=−78.43493…f′(v0​)=−406.37466…⋅13+40.31747…⋅1=−366.05719…v1​=0.78573…
Δv1​=∣0.78573…−1∣=0.21426…Δv1​=0.21426…
v2​=0.64504…:Δv2​=0.14068…
f(v1​)=−101.59366…⋅0.78573…4+20.15873…⋅0.78573…2+3=−23.27682…f′(v1​)=−406.37466…⋅0.78573…3+40.31747…⋅0.78573…=−165.44887…v2​=0.64504…
Δv2​=∣0.64504…−0.78573…∣=0.14068…Δv2​=0.14068…
v3​=0.57039…:Δv3​=0.07465…
f(v2​)=−101.59366…⋅0.64504…4+20.15873…⋅0.64504…2+3=−6.20040…f′(v2​)=−406.37466…⋅0.64504…3+40.31747…⋅0.64504…=−83.05958…v3​=0.57039…
Δv3​=∣0.57039…−0.64504…∣=0.07465…Δv3​=0.07465…
v4​=0.54759…:Δv4​=0.02280…
f(v3​)=−101.59366…⋅0.57039…4+20.15873…⋅0.57039…2+3=−1.19513…f′(v3​)=−406.37466…⋅0.57039…3+40.31747…⋅0.57039…=−52.41610…v4​=0.54759…
Δv4​=∣0.54759…−0.57039…∣=0.02280…Δv4​=0.02280…
v5​=0.54557…:Δv5​=0.00201…
f(v4​)=−101.59366…⋅0.54759…4+20.15873…⋅0.54759…2+3=−0.08990…f′(v4​)=−406.37466…⋅0.54759…3+40.31747…⋅0.54759…=−44.64836…v5​=0.54557…
Δv5​=∣0.54557…−0.54759…∣=0.00201…Δv5​=0.00201…
v6​=0.54556…:Δv6​=0.00001…
f(v5​)=−101.59366…⋅0.54557…4+20.15873…⋅0.54557…2+3=−0.00065…f′(v5​)=−406.37466…⋅0.54557…3+40.31747…⋅0.54557…=−43.99616…v6​=0.54556…
Δv6​=∣0.54556…−0.54557…∣=0.00001…Δv6​=0.00001…
v7​=0.54556…:Δv7​=8.18838E−10
f(v6​)=−101.59366…⋅0.54556…4+20.15873…⋅0.54556…2+3=−3.60218E−8f′(v6​)=−406.37466…⋅0.54556…3+40.31747…⋅0.54556…=−43.99134…v7​=0.54556…
Δv7​=∣0.54556…−0.54556…∣=8.18838E−10Δv7​=8.18838E−10
v≈0.54556…
Apply long division:
−101.59366…v3−55.42562…v2−10.07936…v−5.49891…≈0
Find one solution for −101.59366…v3−55.42562…v2−10.07936…v−5.49891…=0 using Newton-Raphson:v≈−0.54556…
−101.59366…v3−55.42562…v2−10.07936…v−5.49891…=0
Newton-Raphson Approximation Definition
f(v)=−101.59366…v3−55.42562…v2−10.07936…v−5.49891…
Find f′(v):−304.78100…v2−110.85125…v−10.07936…
dvd​(−101.59366…v3−55.42562…v2−10.07936…v−5.49891…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=−dvd​(101.59366…v3)−dvd​(55.42562…v2)−dvd​(10.07936…v)−dvd​(5.49891…)
dvd​(101.59366…v3)=304.78100…v2
dvd​(101.59366…v3)
Take the constant out: (a⋅f)′=a⋅f′=101.59366…dvd​(v3)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=101.59366…⋅3v3−1
Simplify=304.78100…v2
dvd​(55.42562…v2)=110.85125…v
dvd​(55.42562…v2)
Take the constant out: (a⋅f)′=a⋅f′=55.42562…dvd​(v2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=55.42562…⋅2v2−1
Simplify=110.85125…v
dvd​(10.07936…v)=10.07936…
dvd​(10.07936…v)
Take the constant out: (a⋅f)′=a⋅f′=10.07936…dvdv​
Apply the common derivative: dvdv​=1=10.07936…⋅1
Simplify=10.07936…
dvd​(5.49891…)=0
dvd​(5.49891…)
Derivative of a constant: dxd​(a)=0=0
=−304.78100…v2−110.85125…v−10.07936…−0
Simplify=−304.78100…v2−110.85125…v−10.07936…
Let v0​=−1Compute vn+1​ until Δvn+1​<0.000001
v1​=−0.75124…:Δv1​=0.24875…
f(v0​)=−101.59366…(−1)3−55.42562…(−1)2−10.07936…(−1)−5.49891…=50.74849…f′(v0​)=−304.78100…(−1)2−110.85125…(−1)−10.07936…=−204.00911…v1​=−0.75124…
Δv1​=∣−0.75124…−(−1)∣=0.24875…Δv1​=0.24875…
v2​=−0.61091…:Δv2​=0.14032…
f(v1​)=−101.59366…(−0.75124…)3−55.42562…(−0.75124…)2−10.07936…(−0.75124…)−5.49891…=13.86617…f′(v1​)=−304.78100…(−0.75124…)2−110.85125…(−0.75124…)−10.07936…=−98.81153…v2​=−0.61091…
Δv2​=∣−0.61091…−(−0.75124…)∣=0.14032…Δv2​=0.14032…
v3​=−0.55501…:Δv3​=0.05590…
f(v2​)=−101.59366…(−0.61091…)3−55.42562…(−0.61091…)2−10.07936…(−0.61091…)−5.49891…=3.13665…f′(v2​)=−304.78100…(−0.61091…)2−110.85125…(−0.61091…)−10.07936…=−56.10803…v3​=−0.55501…
Δv3​=∣−0.55501…−(−0.61091…)∣=0.05590…Δv3​=0.05590…
v4​=−0.54579…:Δv4​=0.00921…
f(v3​)=−101.59366…(−0.55501…)3−55.42562…(−0.55501…)2−10.07936…(−0.55501…)−5.49891…=0.39093…f′(v3​)=−304.78100…(−0.55501…)2−110.85125…(−0.55501…)−10.07936…=−42.43951…v4​=−0.54579…
Δv4​=∣−0.54579…−(−0.55501…)∣=0.00921…Δv4​=0.00921…
v5​=−0.54556…:Δv5​=0.00023…
f(v4​)=−101.59366…(−0.54579…)3−55.42562…(−0.54579…)2−10.07936…(−0.54579…)−5.49891…=0.00957…f′(v4​)=−304.78100…(−0.54579…)2−110.85125…(−0.54579…)−10.07936…=−40.37008…v5​=−0.54556…
Δv5​=∣−0.54556…−(−0.54579…)∣=0.00023…Δv5​=0.00023…
v6​=−0.54556…:Δv6​=1.54611E−7
f(v5​)=−101.59366…(−0.54556…)3−55.42562…(−0.54556…)2−10.07936…(−0.54556…)−5.49891…=6.23352E−6f′(v5​)=−304.78100…(−0.54556…)2−110.85125…(−0.54556…)−10.07936…=−40.31750…v6​=−0.54556…
Δv6​=∣−0.54556…−(−0.54556…)∣=1.54611E−7Δv6​=1.54611E−7
v≈−0.54556…
Apply long division:v+0.54556…−101.59366…v3−55.42562…v2−10.07936…v−5.49891…​=−101.59366…v2−10.07936…
−101.59366…v2−10.07936…≈0
Find one solution for −101.59366…v2−10.07936…=0 using Newton-Raphson:No Solution for v∈R
−101.59366…v2−10.07936…=0
Newton-Raphson Approximation Definition
f(v)=−101.59366…v2−10.07936…
Find f′(v):−203.18733…v
dvd​(−101.59366…v2−10.07936…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=−dvd​(101.59366…v2)−dvd​(10.07936…)
dvd​(101.59366…v2)=203.18733…v
dvd​(101.59366…v2)
Take the constant out: (a⋅f)′=a⋅f′=101.59366…dvd​(v2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=101.59366…⋅2v2−1
Simplify=203.18733…v
dvd​(10.07936…)=0
dvd​(10.07936…)
Derivative of a constant: dxd​(a)=0=0
=−203.18733…v−0
Simplify=−203.18733…v
Let v0​=−1Compute vn+1​ until Δvn+1​<0.000001
v1​=−0.45039…:Δv1​=0.54960…
f(v0​)=−101.59366…(−1)2−10.07936…=−111.67303…f′(v0​)=−203.18733…(−1)=203.18733…v1​=−0.45039…
Δv1​=∣−0.45039…−(−1)∣=0.54960…Δv1​=0.54960…
v2​=−0.11505…:Δv2​=0.33533…
f(v1​)=−101.59366…(−0.45039…)2−10.07936…=−30.68810…f′(v1​)=−203.18733…(−0.45039…)=91.51429…v2​=−0.11505…
Δv2​=∣−0.11505…−(−0.45039…)∣=0.33533…Δv2​=0.33533…
v3​=0.37361…:Δv3​=0.48867…
f(v2​)=−101.59366…(−0.11505…)2−10.07936…=−11.42427…f′(v2​)=−203.18733…(−0.11505…)=23.37813…v3​=0.37361…
Δv3​=∣0.37361…−(−0.11505…)∣=0.48867…Δv3​=0.48867…
v4​=0.05403…:Δv4​=0.31958…
f(v3​)=−101.59366…⋅0.37361…2−10.07936…=−24.26076…f′(v3​)=−203.18733…⋅0.37361…=−75.91416…v4​=0.05403…
Δv4​=∣0.05403…−0.37361…∣=0.31958…Δv4​=0.31958…
v5​=−0.89102…:Δv5​=0.94505…
f(v4​)=−101.59366…⋅0.05403…2−10.07936…=−10.37600…f′(v4​)=−203.18733…⋅0.05403…=−10.97924…v5​=−0.89102…
Δv5​=∣−0.89102…−0.05403…∣=0.94505…Δv5​=0.94505…
v6​=−0.38983…:Δv6​=0.50118…
f(v5​)=−101.59366…(−0.89102…)2−10.07936…=−90.73642…f′(v5​)=−203.18733…(−0.89102…)=181.04414…v6​=−0.38983…
Δv6​=∣−0.38983…−(−0.89102…)∣=0.50118…Δv6​=0.50118…
v7​=−0.06766…:Δv7​=0.32216…
f(v6​)=−101.59366…(−0.38983…)2−10.07936…=−25.51884…f′(v6​)=−203.18733…(−0.38983…)=79.20991…v7​=−0.06766…
Δv7​=∣−0.06766…−(−0.38983…)∣=0.32216…Δv7​=0.32216…
v8​=0.69923…:Δv8​=0.76690…
f(v7​)=−101.59366…(−0.06766…)2−10.07936…=−10.54458…f′(v7​)=−203.18733…(−0.06766…)=13.74960…v8​=0.69923…
Δv8​=∣0.69923…−(−0.06766…)∣=0.76690…Δv8​=0.76690…
v9​=0.27867…:Δv9​=0.42055…
f(v8​)=−101.59366…⋅0.69923…2−10.07936…=−59.75094…f′(v8​)=−203.18733…⋅0.69923…=−142.07487…v9​=0.27867…
Δv9​=∣0.27867…−0.69923…∣=0.42055…Δv9​=0.42055…
v10​=−0.03867…:Δv10​=0.31734…
f(v9​)=−101.59366…⋅0.27867…2−10.07936…=−17.96890…f′(v9​)=−203.18733…⋅0.27867…=−56.62250…v10​=−0.03867…
Δv10​=∣−0.03867…−0.27867…∣=0.31734…Δv10​=0.31734…
v11​=1.26333…:Δv11​=1.30200…
f(v10​)=−101.59366…(−0.03867…)2−10.07936…=−10.23132…f′(v10​)=−203.18733…(−0.03867…)=7.85811…v11​=1.26333…
Δv11​=∣1.26333…−(−0.03867…)∣=1.30200…Δv11​=1.30200…
Cannot find solution
The solutions arev≈0.54556…,v≈−0.54556…
v≈0.54556…,v≈−0.54556…
Verify Solutions
Find undefined (singularity) points:v=0
Take the denominator(s) of (−2310​v3​​)2−v2 and compare to zero
Solve 2310​v=0:v=0
2310​v=0
Divide both sides by 2310​
2310​v=0
Divide both sides by 2310​2310​2310​v​=2310​0​
Simplify
2310​2310​v​=2310​0​
Simplify 2310​2310​v​:v
2310​2310​v​
Cancel the common factor: 2310​=v
Simplify 2310​0​:0
2310​0​
Apply rule a0​=0: a=0=0
v=0
v=0
v=0
The following points are undefinedv=0
Combine undefined points with solutions:
v≈0.54556…,v≈−0.54556…
Plug the solutions v=0.54556…,v=−0.54556… into 2uv=−8232​⋅3​​
For 2uv=−8232​⋅3​​, subsitute v with
For 2uv=−8232​⋅3​​, subsitute v with 0.54556…2u⋅0.54556…=−8232​3​​
Solve
2u⋅0.54556…=−8232​3​​
Simplify −8232​3​​:−237​3​​
−8232​3​​
Factor the number: 8=23=−23232​3​​
Simplify 23232​​:237​1​
23232​​
Apply exponent rule: xbxa​=xb−a1​=23−32​1​
3−32​=37​
3−32​
Convert element to fraction: 3=33⋅3​=33⋅3​−32​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=33⋅3−2​
3⋅3−2=7
3⋅3−2
Multiply the numbers: 3⋅3=9=9−2
Subtract the numbers: 9−2=7=7
=37​
=237​1​
=−237​3​​
2u⋅0.54556…=−237​3​​
Divide both sides by 2⋅0.54556…
2u⋅0.54556…=−237​3​​
Divide both sides by 2⋅0.54556…2⋅0.54556…2u⋅0.54556…​=2⋅0.54556…−237​3​​​
Simplify
2⋅0.54556…2u⋅0.54556…​=2⋅0.54556…−237​3​​​
Simplify 2⋅0.54556…2u⋅0.54556…​:u
2⋅0.54556…2u⋅0.54556…​
Cancel the common factor: 2=0.54556…u⋅0.54556…​
Cancel the common factor: 0.54556…=u
Simplify
2⋅0.54556…−237​3​​​
Multiply the numbers: 2⋅0.54556…=1.09112…=1.09112…−237​3​​​
Apply the fraction rule: b−a​=−ba​=−1.09112…237​3​​​
Apply the fraction rule: acb​​=c⋅ab​1.09112…237​3​​​=237​⋅1.09112…3​​=−237​⋅1.09112…3​​
237​
237​=22+31​=22+31​
Apply exponent rule: xa+b=xaxb=22⋅231​
Refine
For 2uv=−8232​⋅3​​, subsitute v with
For 2uv=−8232​⋅3​​, subsitute v with −0.54556…2u(−0.54556…)=−8232​3​​
Solve
2u(−0.54556…)=−8232​3​​
Simplify −8232​3​​:−237​3​​
−8232​3​​
Factor the number: 8=23=−23232​3​​
Simplify 23232​​:237​1​
23232​​
Apply exponent rule: xbxa​=xb−a1​=23−32​1​
3−32​=37​
3−32​
Convert element to fraction: 3=33⋅3​=33⋅3​−32​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=33⋅3−2​
3⋅3−2=7
3⋅3−2
Multiply the numbers: 3⋅3=9=9−2
Subtract the numbers: 9−2=7=7
=37​
=237​1​
=−237​3​​
2u(−0.54556…)=−237​3​​
Divide both sides by 2(−0.54556…)
2u(−0.54556…)=−237​3​​
Divide both sides by 2(−0.54556…)2(−0.54556…)2u(−0.54556…)​=2(−0.54556…)−237​3​​​
Simplify
2(−0.54556…)2u(−0.54556…)​=2(−0.54556…)−237​3​​​
Simplify 2(−0.54556…)2u(−0.54556…)​:u
2(−0.54556…)2u(−0.54556…)​
Multiply the numbers: 2(−0.54556…)=−1.09112…=−1.09112…−1.09112…u​
Cancel the common factor: −1.09112…=u
Simplify
2(−0.54556…)−237​3​​​
Remove parentheses: (−a)=−a=−2⋅0.54556…−237​3​​​
Multiply the numbers: 2⋅0.54556…=1.09112…=−1.09112…−237​3​​​
Apply the fraction rule: −b−a​=ba​=1.09112…237​3​​​
Apply the fraction rule: acb​​=c⋅ab​=237​⋅1.09112…3​​
237​
237​=22+31​=22+31​
Apply exponent rule: xa+b=xaxb=22⋅231​
Refine
Verify solutions by plugging them into the original equations
Check the solutions by plugging them into u2−v2=−8232​​
Remove the ones that don't agree with the equation.
Check the solution True
u2−v2=−8232​​
Plug in
Refine
True
Check the solution True
u2−v2=−8232​​
Plug in
Refine
True
Check the solutions by plugging them into 2uv=−83​⋅232​​
Remove the ones that don't agree with the equation.
Check the solution True
2uv=−83​⋅232​​
Plug in
Refine−4.36449…3​⋅232​⋅0.54556…​=−83​⋅232​​
True
Check the solution True
2uv=−83​⋅232​​
Plug in
Refine−4.36449…3​⋅232​⋅0.54556…​=−83​⋅232​​
True
Therefore, the final solutions for u2−v2=−8232​​,2uv=−83​⋅232​​ are
Substitute back w=u+vi
The solutions are
Substitute back w=cos(x)
cos(x)=2232​​​:x=arccos(2232​​​)+2πn,x=2π−arccos(2232​​​)+2πn
cos(x)=2232​​​
Apply trig inverse properties
cos(x)=2232​​​
General solutions for cos(x)=2232​​​cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πnx=arccos(2232​​​)+2πn,x=2π−arccos(2232​​​)+2πn
x=arccos(2232​​​)+2πn,x=2π−arccos(2232​​​)+2πn
cos(x)=−2232​​​:x=arccos(−2232​​​)+2πn,x=−arccos(−2232​​​)+2πn
cos(x)=−2232​​​
Apply trig inverse properties
cos(x)=−2232​​​
General solutions for cos(x)=−2232​​​cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnx=arccos(−2232​​​)+2πn,x=−arccos(−2232​​​)+2πn
x=arccos(−2232​​​)+2πn,x=−arccos(−2232​​​)+2πn
No Solution
Simplify
22=4
Multiply the numbers: 4⋅1.09112…=4.36449…
Multiply by the conjugate 232​232​​
Apply exponent rule: ab⋅ac=ab+c=4.36449…⋅232​+31​
232​+31​=2
232​+31​
Combine the fractions 32​+31​:1
Apply rule ca​±cb​=ca±b​=32+1​
Add the numbers: 2+1=3=33​
Apply rule aa​=1=1
=21
Apply rule a1=a=2
=2⋅4.36449…
Multiply the numbers: 4.36449…⋅2=8.72898…=8.72898…
=8.72898…3​⋅232​​
=8.72898…3​⋅232​​+0.54556…i
NoSolution
No Solution
Simplify
22=4
Multiply the numbers: 4⋅1.09112…=4.36449…
Multiply by the conjugate 232​232​​
Apply exponent rule: ab⋅ac=ab+c=4.36449…⋅232​+31​
232​+31​=2
232​+31​
Combine the fractions 32​+31​:1
Apply rule ca​±cb​=ca±b​=32+1​
Add the numbers: 2+1=3=33​
Apply rule aa​=1=1
=21
Apply rule a1=a=2
=2⋅4.36449…
Multiply the numbers: 4.36449…⋅2=8.72898…=8.72898…
=8.72898…3​⋅232​​
=−8.72898…3​⋅232​​−0.54556…i
NoSolution
No Solution
Simplify
22=4
Multiply the numbers: 4⋅1.09112…=4.36449…
Multiply by the conjugate 232​232​​
Apply exponent rule: ab⋅ac=ab+c=4.36449…⋅232​+31​
232​+31​=2
232​+31​
Combine the fractions 32​+31​:1
Apply rule ca​±cb​=ca±b​=32+1​
Add the numbers: 2+1=3=33​
Apply rule aa​=1=1
=21
Apply rule a1=a=2
=2⋅4.36449…
Multiply the numbers: 4.36449…⋅2=8.72898…=8.72898…
=8.72898…3​⋅232​​
=−8.72898…3​⋅232​​+0.54556…i
NoSolution
No Solution
Simplify
22=4
Multiply the numbers: 4⋅1.09112…=4.36449…
Multiply by the conjugate 232​232​​
Apply exponent rule: ab⋅ac=ab+c=4.36449…⋅232​+31​
232​+31​=2
232​+31​
Combine the fractions 32​+31​:1
Apply rule ca​±cb​=ca±b​=32+1​
Add the numbers: 2+1=3=33​
Apply rule aa​=1=1
=21
Apply rule a1=a=2
=2⋅4.36449…
Multiply the numbers: 4.36449…⋅2=8.72898…=8.72898…
=8.72898…3​⋅232​​
=8.72898…3​⋅232​​−0.54556…i
NoSolution
Combine all the solutionsx=arccos(2232​​​)+2πn,x=2π−arccos(2232​​​)+2πn,x=arccos(−2232​​​)+2πn,x=−arccos(−2232​​​)+2πn
Show solutions in decimal formx=0.88929…+2πn,x=2π−0.88929…+2πn,x=2.25229…+2πn,x=−2.25229…+2πn

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