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Popular Trigonometry >

sin^2(x)+sin^6(x)=3cos^2(2x)

  • Pre Algebra
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Solution

sin2(x)+sin6(x)=3cos2(2x)

Solution

x=0.60819…+2πn,x=π−0.60819…+2πn,x=−0.60819…+2πn,x=π+0.60819…+2πn,x=1.17152…+2πn,x=π−1.17152…+2πn,x=−1.17152…+2πn,x=π+1.17152…+2πn
+1
Degrees
x=34.84715…∘+360∘n,x=145.15284…∘+360∘n,x=−34.84715…∘+360∘n,x=214.84715…∘+360∘n,x=67.12337…∘+360∘n,x=112.87662…∘+360∘n,x=−67.12337…∘+360∘n,x=247.12337…∘+360∘n
Solution steps
sin2(x)+sin6(x)=3cos2(2x)
Subtract 3cos2(2x) from both sidessin2(x)+sin6(x)−3cos2(2x)=0
Rewrite using trig identities
sin2(x)+sin6(x)−3cos2(2x)
Use the Double Angle identity: cos(2x)=1−2sin2(x)=sin2(x)+sin6(x)−3(1−2sin2(x))2
Simplify sin2(x)+sin6(x)−3(1−2sin2(x))2:13sin2(x)+sin6(x)−12sin4(x)−3
sin2(x)+sin6(x)−3(1−2sin2(x))2
(1−2sin2(x))2:1−4sin2(x)+4sin4(x)
Apply Perfect Square Formula: (a−b)2=a2−2ab+b2a=1,b=2sin2(x)
=12−2⋅1⋅2sin2(x)+(2sin2(x))2
Simplify 12−2⋅1⋅2sin2(x)+(2sin2(x))2:1−4sin2(x)+4sin4(x)
12−2⋅1⋅2sin2(x)+(2sin2(x))2
Apply rule 1a=112=1=1−2⋅1⋅2sin2(x)+(2sin2(x))2
2⋅1⋅2sin2(x)=4sin2(x)
2⋅1⋅2sin2(x)
Multiply the numbers: 2⋅1⋅2=4=4sin2(x)
(2sin2(x))2=4sin4(x)
(2sin2(x))2
Apply exponent rule: (a⋅b)n=anbn=22(sin2(x))2
(sin2(x))2:sin4(x)
Apply exponent rule: (ab)c=abc=sin2⋅2(x)
Multiply the numbers: 2⋅2=4=sin4(x)
=22sin4(x)
22=4=4sin4(x)
=1−4sin2(x)+4sin4(x)
=1−4sin2(x)+4sin4(x)
=sin2(x)+sin6(x)−3(1−4sin2(x)+4sin4(x))
Expand −3(1−4sin2(x)+4sin4(x)):−3+12sin2(x)−12sin4(x)
−3(1−4sin2(x)+4sin4(x))
Distribute parentheses=(−3)⋅1+(−3)(−4sin2(x))+(−3)⋅4sin4(x)
Apply minus-plus rules+(−a)=−a,(−a)(−b)=ab=−3⋅1+3⋅4sin2(x)−3⋅4sin4(x)
Simplify −3⋅1+3⋅4sin2(x)−3⋅4sin4(x):−3+12sin2(x)−12sin4(x)
−3⋅1+3⋅4sin2(x)−3⋅4sin4(x)
Multiply the numbers: 3⋅1=3=−3+3⋅4sin2(x)−3⋅4sin4(x)
Multiply the numbers: 3⋅4=12=−3+12sin2(x)−12sin4(x)
=−3+12sin2(x)−12sin4(x)
=sin2(x)+sin6(x)−3+12sin2(x)−12sin4(x)
Simplify sin2(x)+sin6(x)−3+12sin2(x)−12sin4(x):13sin2(x)+sin6(x)−12sin4(x)−3
sin2(x)+sin6(x)−3+12sin2(x)−12sin4(x)
Group like terms=sin2(x)+sin6(x)+12sin2(x)−12sin4(x)−3
Add similar elements: sin2(x)+12sin2(x)=13sin2(x)=13sin2(x)+sin6(x)−12sin4(x)−3
=13sin2(x)+sin6(x)−12sin4(x)−3
=13sin2(x)+sin6(x)−12sin4(x)−3
−3+sin6(x)−12sin4(x)+13sin2(x)=0
Solve by substitution
−3+sin6(x)−12sin4(x)+13sin2(x)=0
Let: sin(x)=u−3+u6−12u4+13u2=0
−3+u6−12u4+13u2=0:u=0.32648…​,u=−0.32648…​,u=0.84887…​,u=−0.84887…​,u=10.82463…​,u=−10.82463…​
−3+u6−12u4+13u2=0
Write in the standard form an​xn+…+a1​x+a0​=0u6−12u4+13u2−3=0
Rewrite the equation with v=u2,v2=u4 and v3=u6v3−12v2+13v−3=0
Solve v3−12v2+13v−3=0:v≈0.32648…,v≈0.84887…,v≈10.82463…
v3−12v2+13v−3=0
Find one solution for v3−12v2+13v−3=0 using Newton-Raphson:v≈0.32648…
v3−12v2+13v−3=0
Newton-Raphson Approximation Definition
f(v)=v3−12v2+13v−3
Find f′(v):3v2−24v+13
dvd​(v3−12v2+13v−3)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dvd​(v3)−dvd​(12v2)+dvd​(13v)−dvd​(3)
dvd​(v3)=3v2
dvd​(v3)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=3v3−1
Simplify=3v2
dvd​(12v2)=24v
dvd​(12v2)
Take the constant out: (a⋅f)′=a⋅f′=12dvd​(v2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=12⋅2v2−1
Simplify=24v
dvd​(13v)=13
dvd​(13v)
Take the constant out: (a⋅f)′=a⋅f′=13dvdv​
Apply the common derivative: dvdv​=1=13⋅1
Simplify=13
dvd​(3)=0
dvd​(3)
Derivative of a constant: dxd​(a)=0=0
=3v2−24v+13−0
Simplify=3v2−24v+13
Let v0​=0Compute vn+1​ until Δvn+1​<0.000001
v1​=0.23076…:Δv1​=0.23076…
f(v0​)=03−12⋅02+13⋅0−3=−3f′(v0​)=3⋅02−24⋅0+13=13v1​=0.23076…
Δv1​=∣0.23076…−0∣=0.23076…Δv1​=0.23076…
v2​=0.31300…:Δv2​=0.08223…
f(v1​)=0.23076…3−12⋅0.23076…2+13⋅0.23076…−3=−0.62676…f′(v1​)=3⋅0.23076…2−24⋅0.23076…+13=7.62130…v2​=0.31300…
Δv2​=∣0.31300…−0.23076…∣=0.08223…Δv2​=0.08223…
v3​=0.32613…:Δv3​=0.01313…
f(v2​)=0.31300…3−12⋅0.31300…2+13⋅0.31300…−3=−0.07591…f′(v2​)=3⋅0.31300…2−24⋅0.31300…+13=5.78173…v3​=0.32613…
Δv3​=∣0.32613…−0.31300…∣=0.01313…Δv3​=0.01313…
v4​=0.32648…:Δv4​=0.00034…
f(v3​)=0.32613…3−12⋅0.32613…2+13⋅0.32613…−3=−0.00190…f′(v3​)=3⋅0.32613…2−24⋅0.32613…+13=5.49177…v4​=0.32648…
Δv4​=∣0.32648…−0.32613…∣=0.00034…Δv4​=0.00034…
v5​=0.32648…:Δv5​=2.41787E−7
f(v4​)=0.32648…3−12⋅0.32648…2+13⋅0.32648…−3=−1.32599E−6f′(v4​)=3⋅0.32648…2−24⋅0.32648…+13=5.48412…v5​=0.32648…
Δv5​=∣0.32648…−0.32648…∣=2.41787E−7Δv5​=2.41787E−7
v≈0.32648…
Apply long division:v−0.32648…v3−12v2+13v−3​=v2−11.67351…v+9.18876…
v2−11.67351…v+9.18876…≈0
Find one solution for v2−11.67351…v+9.18876…=0 using Newton-Raphson:v≈0.84887…
v2−11.67351…v+9.18876…=0
Newton-Raphson Approximation Definition
f(v)=v2−11.67351…v+9.18876…
Find f′(v):2v−11.67351…
dvd​(v2−11.67351…v+9.18876…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dvd​(v2)−dvd​(11.67351…v)+dvd​(9.18876…)
dvd​(v2)=2v
dvd​(v2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=2v2−1
Simplify=2v
dvd​(11.67351…v)=11.67351…
dvd​(11.67351…v)
Take the constant out: (a⋅f)′=a⋅f′=11.67351…dvdv​
Apply the common derivative: dvdv​=1=11.67351…⋅1
Simplify=11.67351…
dvd​(9.18876…)=0
dvd​(9.18876…)
Derivative of a constant: dxd​(a)=0=0
=2v−11.67351…+0
Simplify=2v−11.67351…
Let v0​=1Compute vn+1​ until Δvn+1​<0.000001
v1​=0.84651…:Δv1​=0.15348…
f(v0​)=12−11.67351…⋅1+9.18876…=−1.48474…f′(v0​)=2⋅1−11.67351…=−9.67351…v1​=0.84651…
Δv1​=∣0.84651…−1∣=0.15348…Δv1​=0.15348…
v2​=0.84887…:Δv2​=0.00236…
f(v1​)=0.84651…2−11.67351…⋅0.84651…+9.18876…=0.02355…f′(v1​)=2⋅0.84651…−11.67351…=−9.98048…v2​=0.84887…
Δv2​=∣0.84887…−0.84651…∣=0.00236…Δv2​=0.00236…
v3​=0.84887…:Δv3​=5.58503E−7
f(v2​)=0.84887…2−11.67351…⋅0.84887…+9.18876…=5.5715E−6f′(v2​)=2⋅0.84887…−11.67351…=−9.97576…v3​=0.84887…
Δv3​=∣0.84887…−0.84887…∣=5.58503E−7Δv3​=5.58503E−7
v≈0.84887…
Apply long division:v−0.84887…v2−11.67351…v+9.18876…​=v−10.82463…
v−10.82463…≈0
v≈10.82463…
The solutions arev≈0.32648…,v≈0.84887…,v≈10.82463…
v≈0.32648…,v≈0.84887…,v≈10.82463…
Substitute back v=u2,solve for u
Solve u2=0.32648…:u=0.32648…​,u=−0.32648…​
u2=0.32648…
For x2=f(a) the solutions are x=f(a)​,−f(a)​
u=0.32648…​,u=−0.32648…​
Solve u2=0.84887…:u=0.84887…​,u=−0.84887…​
u2=0.84887…
For x2=f(a) the solutions are x=f(a)​,−f(a)​
u=0.84887…​,u=−0.84887…​
Solve u2=10.82463…:u=10.82463…​,u=−10.82463…​
u2=10.82463…
For x2=f(a) the solutions are x=f(a)​,−f(a)​
u=10.82463…​,u=−10.82463…​
The solutions are
u=0.32648…​,u=−0.32648…​,u=0.84887…​,u=−0.84887…​,u=10.82463…​,u=−10.82463…​
Substitute back u=sin(x)sin(x)=0.32648…​,sin(x)=−0.32648…​,sin(x)=0.84887…​,sin(x)=−0.84887…​,sin(x)=10.82463…​,sin(x)=−10.82463…​
sin(x)=0.32648…​,sin(x)=−0.32648…​,sin(x)=0.84887…​,sin(x)=−0.84887…​,sin(x)=10.82463…​,sin(x)=−10.82463…​
sin(x)=0.32648…​:x=arcsin(0.32648…​)+2πn,x=π−arcsin(0.32648…​)+2πn
sin(x)=0.32648…​
Apply trig inverse properties
sin(x)=0.32648…​
General solutions for sin(x)=0.32648…​sin(x)=a⇒x=arcsin(a)+2πn,x=π−arcsin(a)+2πnx=arcsin(0.32648…​)+2πn,x=π−arcsin(0.32648…​)+2πn
x=arcsin(0.32648…​)+2πn,x=π−arcsin(0.32648…​)+2πn
sin(x)=−0.32648…​:x=arcsin(−0.32648…​)+2πn,x=π+arcsin(0.32648…​)+2πn
sin(x)=−0.32648…​
Apply trig inverse properties
sin(x)=−0.32648…​
General solutions for sin(x)=−0.32648…​sin(x)=−a⇒x=arcsin(−a)+2πn,x=π+arcsin(a)+2πnx=arcsin(−0.32648…​)+2πn,x=π+arcsin(0.32648…​)+2πn
x=arcsin(−0.32648…​)+2πn,x=π+arcsin(0.32648…​)+2πn
sin(x)=0.84887…​:x=arcsin(0.84887…​)+2πn,x=π−arcsin(0.84887…​)+2πn
sin(x)=0.84887…​
Apply trig inverse properties
sin(x)=0.84887…​
General solutions for sin(x)=0.84887…​sin(x)=a⇒x=arcsin(a)+2πn,x=π−arcsin(a)+2πnx=arcsin(0.84887…​)+2πn,x=π−arcsin(0.84887…​)+2πn
x=arcsin(0.84887…​)+2πn,x=π−arcsin(0.84887…​)+2πn
sin(x)=−0.84887…​:x=arcsin(−0.84887…​)+2πn,x=π+arcsin(0.84887…​)+2πn
sin(x)=−0.84887…​
Apply trig inverse properties
sin(x)=−0.84887…​
General solutions for sin(x)=−0.84887…​sin(x)=−a⇒x=arcsin(−a)+2πn,x=π+arcsin(a)+2πnx=arcsin(−0.84887…​)+2πn,x=π+arcsin(0.84887…​)+2πn
x=arcsin(−0.84887…​)+2πn,x=π+arcsin(0.84887…​)+2πn
sin(x)=10.82463…​:No Solution
sin(x)=10.82463…​
−1≤sin(x)≤1NoSolution
sin(x)=−10.82463…​:No Solution
sin(x)=−10.82463…​
−1≤sin(x)≤1NoSolution
Combine all the solutionsx=arcsin(0.32648…​)+2πn,x=π−arcsin(0.32648…​)+2πn,x=arcsin(−0.32648…​)+2πn,x=π+arcsin(0.32648…​)+2πn,x=arcsin(0.84887…​)+2πn,x=π−arcsin(0.84887…​)+2πn,x=arcsin(−0.84887…​)+2πn,x=π+arcsin(0.84887…​)+2πn
Show solutions in decimal formx=0.60819…+2πn,x=π−0.60819…+2πn,x=−0.60819…+2πn,x=π+0.60819…+2πn,x=1.17152…+2πn,x=π−1.17152…+2πn,x=−1.17152…+2πn,x=π+1.17152…+2πn

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