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Popular Trigonometry >

cos(3x)cos(x)=2cos(2x)cos(x)

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Solution

cos(3x)cos(x)=2cos(2x)cos(x)

Solution

x=2π​+2πn,x=23π​+2πn,x=1.04719…+2πn,x=2π−1.04719…+2πn,x=2.46670…+2πn,x=−2.46670…+2πn
+1
Degrees
x=90∘+360∘n,x=270∘+360∘n,x=60.00000…∘+360∘n,x=299.99999…∘+360∘n,x=141.33171…∘+360∘n,x=−141.33171…∘+360∘n
Solution steps
cos(3x)cos(x)=2cos(2x)cos(x)
Subtract 2cos(2x)cos(x) from both sidescos(3x)cos(x)−2cos(2x)cos(x)=0
Rewrite using trig identities
cos(3x)cos(x)−2cos(2x)cos(x)
Use the Double Angle identity: cos(2x)=2cos2(x)−1=cos(3x)cos(x)−2(2cos2(x)−1)cos(x)
cos(3x)cos(x)−(−1+2cos2(x))⋅2cos(x)=0
Factor cos(3x)cos(x)−(−1+2cos2(x))⋅2cos(x):cos(x)(cos(3x)−2(2​cos(x)+1)(2​cos(x)−1))
cos(3x)cos(x)−(−1+2cos2(x))⋅2cos(x)
Factor out common term cos(x)=cos(x)(cos(3x)−2(−1+cos2(x)⋅2))
Factor cos(3x)−2(2cos2(x)−1):cos(3x)−2(2​cos(x)+1)(2​cos(x)−1)
cos(3x)−2(−1+cos2(x)⋅2)
Factor −1+cos2(x)⋅2:(2​cos(x)+1)(2​cos(x)−1)
−1+cos2(x)⋅2
Rewrite 2cos2(x)−1 as (2​cos(x))2−12
2cos2(x)−1
Apply radical rule: a=(a​)22=(2​)2=(2​)2cos2(x)−1
Rewrite 1 as 12=(2​)2cos2(x)−12
Apply exponent rule: ambm=(ab)m(2​)2cos2(x)=(2​cos(x))2=(2​cos(x))2−12
=(2​cos(x))2−12
Apply Difference of Two Squares Formula: x2−y2=(x+y)(x−y)(2​cos(x))2−12=(2​cos(x)+1)(2​cos(x)−1)=(2​cos(x)+1)(2​cos(x)−1)
=cos(3x)−2(2​cos(x)+1)(2​cos(x)−1)
=cos(x)(cos(3x)−2(2​cos(x)+1)(2​cos(x)−1))
cos(x)(cos(3x)−2(2​cos(x)+1)(2​cos(x)−1))=0
Solving each part separatelycos(x)=0orcos(3x)−2(2​cos(x)+1)(2​cos(x)−1)=0
cos(x)=0:x=2π​+2πn,x=23π​+2πn
cos(x)=0
General solutions for cos(x)=0
cos(x) periodicity table with 2πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​cos(x)123​​22​​21​0−21​−22​​−23​​​xπ67π​45π​34π​23π​35π​47π​611π​​cos(x)−1−23​​−22​​−21​021​22​​23​​​​
x=2π​+2πn,x=23π​+2πn
x=2π​+2πn,x=23π​+2πn
cos(3x)−2(2​cos(x)+1)(2​cos(x)−1)=0:x=arccos(0.50000…)+2πn,x=2π−arccos(0.50000…)+2πn,x=arccos(−0.78077…)+2πn,x=−arccos(−0.78077…)+2πn
cos(3x)−2(2​cos(x)+1)(2​cos(x)−1)=0
Rewrite using trig identities
cos(3x)−(−1+cos(x)2​)(1+cos(x)2​)⋅2
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)
=4cos3(x)−3cos(x)−2(−1+2​cos(x))(1+2​cos(x))
−3cos(x)+4cos3(x)−(−1+cos(x)2​)(1+cos(x)2​)⋅2=0
Solve by substitution
−3cos(x)+4cos3(x)−(−1+cos(x)2​)(1+cos(x)2​)⋅2=0
Let: cos(x)=u−3u+4u3−(−1+u2​)(1+u2​)⋅2=0
−3u+4u3−(−1+u2​)(1+u2​)⋅2=0:u≈1.28077…,u≈0.50000…,u≈−0.78077…
−3u+4u3−(−1+u2​)(1+u2​)⋅2=0
Expand −3u+4u3−(−1+u2​)(1+u2​)⋅2:−3u+4u3−4u2+2
−3u+4u3−(−1+u2​)(1+u2​)⋅2
=−3u+4u3−2(−1+2​u)(1+2​u)
Expand −(−1+u2​)(1+u2​)⋅2:−4u2+2
Expand (−1+u2​)(1+u2​):2u2−1
(−1+u2​)(1+u2​)
Apply Difference of Two Squares Formula: (a−b)(a+b)=a2−b2a=u2​,b=1=(u2​)2−12
Simplify (u2​)2−12:2u2−1
(u2​)2−12
Apply rule 1a=112=1=(2​u)2−1
(u2​)2=2u2
(u2​)2
Apply exponent rule: (a⋅b)n=anbn=(2​)2u2
(2​)2:2
Apply radical rule: a​=a21​=(221​)2
Apply exponent rule: (ab)c=abc=221​⋅2
21​⋅2=1
21​⋅2
Multiply fractions: a⋅cb​=ca⋅b​=21⋅2​
Cancel the common factor: 2=1
=2
=u2⋅2
=2u2−1
=2u2−1
=−2(2u2−1)
Expand −2(2u2−1):−4u2+2
−2(2u2−1)
Apply the distributive law: a(b−c)=ab−aca=−2,b=2u2,c=1=−2⋅2u2−(−2)⋅1
Apply minus-plus rules−(−a)=a=−2⋅2u2+2⋅1
Simplify −2⋅2u2+2⋅1:−4u2+2
−2⋅2u2+2⋅1
Multiply the numbers: 2⋅2=4=−4u2+2⋅1
Multiply the numbers: 2⋅1=2=−4u2+2
=−4u2+2
=−4u2+2
=−3u+4u3−4u2+2
−3u+4u3−4u2+2=0
Write in the standard form an​xn+…+a1​x+a0​=04u3−4u2−3u+2=0
Find one solution for 4u3−4u2−3u+2=0 using Newton-Raphson:u≈1.28077…
4u3−4u2−3u+2=0
Newton-Raphson Approximation Definition
f(u)=4u3−4u2−3u+2
Find f′(u):12u2−8u−3
dud​(4u3−4u2−3u+2)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(4u3)−dud​(4u2)−dud​(3u)+dud​(2)
dud​(4u3)=12u2
dud​(4u3)
Take the constant out: (a⋅f)′=a⋅f′=4dud​(u3)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=4⋅3u3−1
Simplify=12u2
dud​(4u2)=8u
dud​(4u2)
Take the constant out: (a⋅f)′=a⋅f′=4dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=4⋅2u2−1
Simplify=8u
dud​(3u)=3
dud​(3u)
Take the constant out: (a⋅f)′=a⋅f′=3dudu​
Apply the common derivative: dudu​=1=3⋅1
Simplify=3
dud​(2)=0
dud​(2)
Derivative of a constant: dxd​(a)=0=0
=12u2−8u−3+0
Simplify=12u2−8u−3
Let u0​=1Compute un+1​ until Δun+1​<0.000001
u1​=2:Δu1​=1
f(u0​)=4⋅13−4⋅12−3⋅1+2=−1f′(u0​)=12⋅12−8⋅1−3=1u1​=2
Δu1​=∣2−1∣=1Δu1​=1
u2​=1.58620…:Δu2​=0.41379…
f(u1​)=4⋅23−4⋅22−3⋅2+2=12f′(u1​)=12⋅22−8⋅2−3=29u2​=1.58620…
Δu2​=∣1.58620…−2∣=0.41379…Δu2​=0.41379…
u3​=1.36962…:Δu3​=0.21658…
f(u2​)=4⋅1.58620…3−4⋅1.58620…2−3⋅1.58620…+2=3.14108…f′(u2​)=12⋅1.58620…2−8⋅1.58620…−3=14.50297…u3​=1.36962…
Δu3​=∣1.36962…−1.58620…∣=0.21658…Δu3​=0.21658…
u4​=1.29192…:Δu4​=0.07769…
f(u3​)=4⋅1.36962…3−4⋅1.36962…2−3⋅1.36962…+2=0.66459…f′(u3​)=12⋅1.36962…2−8⋅1.36962…−3=8.55345…u4​=1.29192…
Δu4​=∣1.29192…−1.36962…∣=0.07769…Δu4​=0.07769…
u5​=1.28098…:Δu5​=0.01093…
f(u4​)=4⋅1.29192…3−4⋅1.29192…2−3⋅1.29192…+2=0.07319…f′(u4​)=12⋅1.29192…2−8⋅1.29192…−3=6.69344…u5​=1.28098…
Δu5​=∣1.28098…−1.29192…∣=0.01093…Δu5​=0.01093…
u6​=1.28077…:Δu6​=0.00021…
f(u5​)=4⋅1.28098…3−4⋅1.28098…2−3⋅1.28098…+2=0.00137…f′(u5​)=12⋅1.28098…2−8⋅1.28098…−3=6.44328…u6​=1.28077…
Δu6​=∣1.28077…−1.28098…∣=0.00021…Δu6​=0.00021…
u7​=1.28077…:Δu7​=7.99005E−8
f(u6​)=4⋅1.28077…3−4⋅1.28077…2−3⋅1.28077…+2=5.14435E−7f′(u6​)=12⋅1.28077…2−8⋅1.28077…−3=6.43844…u7​=1.28077…
Δu7​=∣1.28077…−1.28077…∣=7.99005E−8Δu7​=7.99005E−8
u≈1.28077…
Apply long division:u−1.28077…4u3−4u2−3u+2​=4u2+1.12310…u−1.56155…
4u2+1.12310…u−1.56155…≈0
Find one solution for 4u2+1.12310…u−1.56155…=0 using Newton-Raphson:u≈0.50000…
4u2+1.12310…u−1.56155…=0
Newton-Raphson Approximation Definition
f(u)=4u2+1.12310…u−1.56155…
Find f′(u):8u+1.12310…
dud​(4u2+1.12310…u−1.56155…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(4u2)+dud​(1.12310…u)−dud​(1.56155…)
dud​(4u2)=8u
dud​(4u2)
Take the constant out: (a⋅f)′=a⋅f′=4dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=4⋅2u2−1
Simplify=8u
dud​(1.12310…u)=1.12310…
dud​(1.12310…u)
Take the constant out: (a⋅f)′=a⋅f′=1.12310…dudu​
Apply the common derivative: dudu​=1=1.12310…⋅1
Simplify=1.12310…
dud​(1.56155…)=0
dud​(1.56155…)
Derivative of a constant: dxd​(a)=0=0
=8u+1.12310…−0
Simplify=8u+1.12310…
Let u0​=1Compute un+1​ until Δun+1​<0.000001
u1​=0.60961…:Δu1​=0.39038…
f(u0​)=4⋅12+1.12310…⋅1−1.56155…=3.56155…f′(u0​)=8⋅1+1.12310…=9.12310…u1​=0.60961…
Δu1​=∣0.60961…−1∣=0.39038…Δu1​=0.39038…
u2​=0.50800…:Δu2​=0.10160…
f(u1​)=4⋅0.60961…2+1.12310…⋅0.60961…−1.56155…=0.60961…f′(u1​)=8⋅0.60961…+1.12310…=6u2​=0.50800…
Δu2​=∣0.50800…−0.60961…∣=0.10160…Δu2​=0.10160…
u3​=0.50004…:Δu3​=0.00796…
f(u2​)=4⋅0.50800…2+1.12310…⋅0.50800…−1.56155…=0.04129…f′(u2​)=8⋅0.50800…+1.12310…=5.18718…u3​=0.50004…
Δu3​=∣0.50004…−0.50800…∣=0.00796…Δu3​=0.00796…
u4​=0.50000…:Δu4​=0.00004…
f(u3​)=4⋅0.50004…2+1.12310…⋅0.50004…−1.56155…=0.00025…f′(u3​)=8⋅0.50004…+1.12310…=5.12350…u4​=0.50000…
Δu4​=∣0.50000…−0.50004…∣=0.00004…Δu4​=0.00004…
u5​=0.5:Δu5​=1.91092E−9
f(u4​)=4⋅0.50000…2+1.12310…⋅0.50000…−1.56155…=9.78986E−9f′(u4​)=8⋅0.50000…+1.12310…=5.12310…u5​=0.5
Δu5​=∣0.5−0.50000…∣=1.91092E−9Δu5​=1.91092E−9
u≈0.50000…
Apply long division:u−0.54u2+1.12310…u−1.56155…​=4u+3.12310…
4u+3.12310…≈0
u≈−0.78077…
The solutions areu≈1.28077…,u≈0.50000…,u≈−0.78077…
Substitute back u=cos(x)cos(x)≈1.28077…,cos(x)≈0.50000…,cos(x)≈−0.78077…
cos(x)≈1.28077…,cos(x)≈0.50000…,cos(x)≈−0.78077…
cos(x)=1.28077…:No Solution
cos(x)=1.28077…
−1≤cos(x)≤1NoSolution
cos(x)=0.50000…:x=arccos(0.50000…)+2πn,x=2π−arccos(0.50000…)+2πn
cos(x)=0.50000…
Apply trig inverse properties
cos(x)=0.50000…
General solutions for cos(x)=0.50000…cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πnx=arccos(0.50000…)+2πn,x=2π−arccos(0.50000…)+2πn
x=arccos(0.50000…)+2πn,x=2π−arccos(0.50000…)+2πn
cos(x)=−0.78077…:x=arccos(−0.78077…)+2πn,x=−arccos(−0.78077…)+2πn
cos(x)=−0.78077…
Apply trig inverse properties
cos(x)=−0.78077…
General solutions for cos(x)=−0.78077…cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnx=arccos(−0.78077…)+2πn,x=−arccos(−0.78077…)+2πn
x=arccos(−0.78077…)+2πn,x=−arccos(−0.78077…)+2πn
Combine all the solutionsx=arccos(0.50000…)+2πn,x=2π−arccos(0.50000…)+2πn,x=arccos(−0.78077…)+2πn,x=−arccos(−0.78077…)+2πn
Combine all the solutionsx=2π​+2πn,x=23π​+2πn,x=arccos(0.50000…)+2πn,x=2π−arccos(0.50000…)+2πn,x=arccos(−0.78077…)+2πn,x=−arccos(−0.78077…)+2πn
Show solutions in decimal formx=2π​+2πn,x=23π​+2πn,x=1.04719…+2πn,x=2π−1.04719…+2πn,x=2.46670…+2πn,x=−2.46670…+2πn

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