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Popular Trigonometry >

cos(4θ)+cos(2θ)=cos(θ)

  • Pre Algebra
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Solution

cos(4θ)+cos(2θ)=cos(θ)

Solution

θ=2π​+2πn,θ=23π​+2πn,θ=9π​+32πn​,θ=95π​+32πn​
+1
Degrees
θ=90∘+360∘n,θ=270∘+360∘n,θ=20∘+120∘n,θ=100∘+120∘n
Solution steps
cos(4θ)+cos(2θ)=cos(θ)
Subtract cos(θ) from both sidescos(4θ)+cos(2θ)−cos(θ)=0
Rewrite using trig identities
cos(2θ)+cos(4θ)−cos(θ)
Use the Sum to Product identity: cos(s)+cos(t)=2cos(2s+t​)cos(2s−t​)=−cos(θ)+2cos(22θ+4θ​)cos(22θ−4θ​)
2cos(22θ+4θ​)cos(22θ−4θ​)=2cos(θ)cos(3θ)
2cos(22θ+4θ​)cos(22θ−4θ​)
22θ+4θ​=3θ
22θ+4θ​
Add similar elements: 2θ+4θ=6θ=26θ​
Divide the numbers: 26​=3=3θ
=2cos(3θ)cos(22θ−4θ​)
22θ−4θ​=−θ
22θ−4θ​
Add similar elements: 2θ−4θ=−2θ=2−2θ​
Apply the fraction rule: b−a​=−ba​=−22θ​
Divide the numbers: 22​=1=−θ
=2cos(3θ)cos(−θ)
Use the negative angle identity: cos(−x)=cos(x)=2cos(θ)cos(3θ)
=−cos(θ)+2cos(θ)cos(3θ)
−cos(θ)+2cos(3θ)cos(θ)=0
Factor −cos(θ)+2cos(3θ)cos(θ):cos(θ)(2cos(3θ)−1)
−cos(θ)+2cos(3θ)cos(θ)
Factor out common term cos(θ)=cos(θ)(−1+2cos(3θ))
cos(θ)(2cos(3θ)−1)=0
Solving each part separatelycos(θ)=0or2cos(3θ)−1=0
cos(θ)=0:θ=2π​+2πn,θ=23π​+2πn
cos(θ)=0
General solutions for cos(θ)=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​​​​
θ=2π​+2πn,θ=23π​+2πn
θ=2π​+2πn,θ=23π​+2πn
2cos(3θ)−1=0:θ=9π​+32πn​,θ=95π​+32πn​
2cos(3θ)−1=0
Move 1to the right side
2cos(3θ)−1=0
Add 1 to both sides2cos(3θ)−1+1=0+1
Simplify2cos(3θ)=1
2cos(3θ)=1
Divide both sides by 2
2cos(3θ)=1
Divide both sides by 222cos(3θ)​=21​
Simplifycos(3θ)=21​
cos(3θ)=21​
General solutions for cos(3θ)=21​
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​​​​
3θ=3π​+2πn,3θ=35π​+2πn
3θ=3π​+2πn,3θ=35π​+2πn
Solve 3θ=3π​+2πn:θ=9π​+32πn​
3θ=3π​+2πn
Divide both sides by 3
3θ=3π​+2πn
Divide both sides by 333θ​=33π​​+32πn​
Simplify
33θ​=33π​​+32πn​
Simplify 33θ​:θ
33θ​
Divide the numbers: 33​=1=θ
Simplify 33π​​+32πn​:9π​+32πn​
33π​​+32πn​
33π​​=9π​
33π​​
Apply the fraction rule: acb​​=c⋅ab​=3⋅3π​
Multiply the numbers: 3⋅3=9=9π​
=9π​+32πn​
θ=9π​+32πn​
θ=9π​+32πn​
θ=9π​+32πn​
Solve 3θ=35π​+2πn:θ=95π​+32πn​
3θ=35π​+2πn
Divide both sides by 3
3θ=35π​+2πn
Divide both sides by 333θ​=335π​​+32πn​
Simplify
33θ​=335π​​+32πn​
Simplify 33θ​:θ
33θ​
Divide the numbers: 33​=1=θ
Simplify 335π​​+32πn​:95π​+32πn​
335π​​+32πn​
335π​​=95π​
335π​​
Apply the fraction rule: acb​​=c⋅ab​=3⋅35π​
Multiply the numbers: 3⋅3=9=95π​
=95π​+32πn​
θ=95π​+32πn​
θ=95π​+32πn​
θ=95π​+32πn​
θ=9π​+32πn​,θ=95π​+32πn​
Combine all the solutionsθ=2π​+2πn,θ=23π​+2πn,θ=9π​+32πn​,θ=95π​+32πn​

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