# Solve `2 - 2*sin^2x = cos x` for x in [0, 360]

lemjay | Certified Educator

`2-2sin^2 x = cos x`

Factor out 2 at the left side of the equation.

`2 ( 1 - sin^2x) = cos x`

From the pythagorean identity `sin^2x + cos^2x = 1` ,                     replace   `1-sin^2x`  with  ` cos^2x` .

`2cos^2x = cos x`

Then, substract both sides by cos x.

`2cos^2x - cos x = 0`

Factor left side.

`cos x (2cosx - 1) = 0`

Set each factor to zero and solve for x.

`cos x = 0 `                    and        `2cos x - 1 = 0`

`2cos x = 1`

` cos x = 1/2`

Refer to Unit circle Chart or Table of Trigonometric Function for Special Angles to determine the value of angle x.

Base on this, the values of x are

`cos x = 0`      ====>  x = 90 and 270 degrees

`cos x = 1/2`      ====>  x = 60 and 300 degrees

Hence, the values of x in the interval [0,360] are 60, 90, 270 and 300 degrees.

justaguide | Certified Educator

The equation `2 - 2*sin^2 x = cos x` has to be solved for x in [0, 360]

`2 - 2*sin^2 x = cos x`

=> `2 - 2*(1 - cos^2x) = cos x`

=> `2 - 2 + 2*cos^2x = cos x`

=> `2*cos^2x - cos x = 0`

=> `cos x(2*cos x - 1) = 0`

=> cos x = 0 and cos x = 1/2

x = 90, x = 270, x = 60 and x = 300

The solution of the equation is {60, 90, 270, 300}