Slove cos^2x=-cosx+sin^2x for 0°≤ x ≤ 360°?

To solve the equation cos^2x = -cosx + sin^2x, we can use the Pythagorean identity sin^2x + cos^2x = 1.

Substitute sin^2x = 1 - cos^2x into the original equation:

cos^2x = -cosx + 1 - cos^2x
2cos^2x - cosx - 1 = 0

Now, let's solve this quadratic equation for cosx:

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

Setting each factor to zero gives:

2cosx + 1 = 0
cosx = -1/2

cosx - 1 = 0
cosx = 1

Therefore, the solutions for 0° ≤ x ≤ 360° are:

x = 120°, 240°, 0°, 360°