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Popular Trigonometry >

2cos^4(x)-2cos^2(x)=4cos^2(x)-1

  • Pre Algebra
  • Algebra
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Solution

2cos4(x)−2cos2(x)=4cos2(x)−1

Solution

x=1.13640…+2πn,x=2π−1.13640…+2πn,x=2.00519…+2πn,x=−2.00519…+2πn
+1
Degrees
x=65.11101…∘+360∘n,x=294.88898…∘+360∘n,x=114.88898…∘+360∘n,x=−114.88898…∘+360∘n
Solution steps
2cos4(x)−2cos2(x)=4cos2(x)−1
Solve by substitution
2cos4(x)−2cos2(x)=4cos2(x)−1
Let: cos(x)=u2u4−2u2=4u2−1
2u4−2u2=4u2−1:u=23+7​​​,u=−23+7​​​,u=23−7​​​,u=−23−7​​​
2u4−2u2=4u2−1
Move 1to the left side
2u4−2u2=4u2−1
Add 1 to both sides2u4−2u2+1=4u2−1+1
Simplify2u4−2u2+1=4u2
2u4−2u2+1=4u2
Move 4u2to the left side
2u4−2u2+1=4u2
Subtract 4u2 from both sides2u4−2u2+1−4u2=4u2−4u2
Simplify2u4−6u2+1=0
2u4−6u2+1=0
Rewrite the equation with v=u2 and v2=u42v2−6v+1=0
Solve 2v2−6v+1=0:v=23+7​​,v=23−7​​
2v2−6v+1=0
Solve with the quadratic formula
2v2−6v+1=0
Quadratic Equation Formula:
For a=2,b=−6,c=1v1,2​=2⋅2−(−6)±(−6)2−4⋅2⋅1​​
v1,2​=2⋅2−(−6)±(−6)2−4⋅2⋅1​​
(−6)2−4⋅2⋅1​=27​
(−6)2−4⋅2⋅1​
Apply exponent rule: (−a)n=an,if n is even(−6)2=62=62−4⋅2⋅1​
Multiply the numbers: 4⋅2⋅1=8=62−8​
62=36=36−8​
Subtract the numbers: 36−8=28=28​
Prime factorization of 28:22⋅7
28
28divides by 228=14⋅2=2⋅14
14divides by 214=7⋅2=2⋅2⋅7
2,7 are all prime numbers, therefore no further factorization is possible=2⋅2⋅7
=22⋅7
=22⋅7​
Apply radical rule: =7​22​
Apply radical rule: 22​=2=27​
v1,2​=2⋅2−(−6)±27​​
Separate the solutionsv1​=2⋅2−(−6)+27​​,v2​=2⋅2−(−6)−27​​
v=2⋅2−(−6)+27​​:23+7​​
2⋅2−(−6)+27​​
Apply rule −(−a)=a=2⋅26+27​​
Multiply the numbers: 2⋅2=4=46+27​​
Factor 6+27​:2(3+7​)
6+27​
Rewrite as=2⋅3+27​
Factor out common term 2=2(3+7​)
=42(3+7​)​
Cancel the common factor: 2=23+7​​
v=2⋅2−(−6)−27​​:23−7​​
2⋅2−(−6)−27​​
Apply rule −(−a)=a=2⋅26−27​​
Multiply the numbers: 2⋅2=4=46−27​​
Factor 6−27​:2(3−7​)
6−27​
Rewrite as=2⋅3−27​
Factor out common term 2=2(3−7​)
=42(3−7​)​
Cancel the common factor: 2=23−7​​
The solutions to the quadratic equation are:v=23+7​​,v=23−7​​
v=23+7​​,v=23−7​​
Substitute back v=u2,solve for u
Solve u2=23+7​​:u=23+7​​​,u=−23+7​​​
u2=23+7​​
For x2=f(a) the solutions are x=f(a)​,−f(a)​
u=23+7​​​,u=−23+7​​​
Solve u2=23−7​​:u=23−7​​​,u=−23−7​​​
u2=23−7​​
For x2=f(a) the solutions are x=f(a)​,−f(a)​
u=23−7​​​,u=−23−7​​​
The solutions are
u=23+7​​​,u=−23+7​​​,u=23−7​​​,u=−23−7​​​
Substitute back u=cos(x)cos(x)=23+7​​​,cos(x)=−23+7​​​,cos(x)=23−7​​​,cos(x)=−23−7​​​
cos(x)=23+7​​​,cos(x)=−23+7​​​,cos(x)=23−7​​​,cos(x)=−23−7​​​
cos(x)=23+7​​​:No Solution
cos(x)=23+7​​​
−1≤cos(x)≤1NoSolution
cos(x)=−23+7​​​:No Solution
cos(x)=−23+7​​​
−1≤cos(x)≤1NoSolution
cos(x)=23−7​​​:x=arccos​23−7​​​​+2πn,x=2π−arccos​23−7​​​​+2πn
cos(x)=23−7​​​
Apply trig inverse properties
cos(x)=23−7​​​
General solutions for cos(x)=23−7​​​cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πnx=arccos​23−7​​​​+2πn,x=2π−arccos​23−7​​​​+2πn
x=arccos​23−7​​​​+2πn,x=2π−arccos​23−7​​​​+2πn
cos(x)=−23−7​​​:x=arccos​−23−7​​​​+2πn,x=−arccos​−23−7​​​​+2πn
cos(x)=−23−7​​​
Apply trig inverse properties
cos(x)=−23−7​​​
General solutions for cos(x)=−23−7​​​cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnx=arccos​−23−7​​​​+2πn,x=−arccos​−23−7​​​​+2πn
x=arccos​−23−7​​​​+2πn,x=−arccos​−23−7​​​​+2πn
Combine all the solutionsx=arccos​23−7​​​​+2πn,x=2π−arccos​23−7​​​​+2πn,x=arccos​−23−7​​​​+2πn,x=−arccos​−23−7​​​​+2πn
Show solutions in decimal formx=1.13640…+2πn,x=2π−1.13640…+2πn,x=2.00519…+2πn,x=−2.00519…+2πn

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