Find all solutions to the equation. 2sin^2(x)+cosx=1 Write your answer in radians in terms of pi , and use the "or" button as necessary

Question

Find all solutions to the equation.

2sin²x+cosx=1

Write your answer in radians in terms of , and use the "or" button as necessary.

Answer 1

To solve this, we need to convert the sin^2 term in terms of cos.

sin^2 = 1 - cos^2

Thus, we have 2(1 - cos^2(x)) + cos(x) = 1, or

-2cos^2(x) + cos(x) + 1 = 0, or

(2 cos(x) + 1)(-cos(x) + 1) = 0

cos(x) = 1 or cos(x) = -1/2

If we want solutions only in [0, 2π), we have, from cos x = 1, 0; from cos x = -1/2, we have 2π/3 and 4π/3

0 or 2π/3 or 4π/3

Answer 2

2sin²x+cosx=1

2sin²x+cosx-1=0

-1 + cosx + 2(1-cos²x)=0 , using the sin²x+cos²x=1

-1 + cosx + 2 - 2cos²x = 0

-2cos²x + cosx + 1 = 0

let cosx=u

-2u² + u + 1 = 0

2u² - u - 1 = 0

2u² - 2u + u -1 = 0

2u(u-1) + 1(u-1) = 0

(u-1)(2u+1)=0

u=1, u=-1/2

therefore cosx=1 and cosx=-1/2

For cosx=1, x = 2pi/3 + 2pi*n, n∈Z or x = 4pi/3 + 2pi*n, n∈Z

For cosx=-1/2, x = 2pi*n, n∈Z

Combine Solution is x = 2pi/3 + 2pi*n, n∈Z or x = 4pi/3 + 2pi*n, n∈Z , x = 2pi*n, n∈Z

Answer 3

If you make the substitution sin²x = 1- cos²x, you will get a quadratic equation in cos x which can be solved by the quadratic formula. Then you may need to remember that cos is periodic to get "all" the solutions.