Solve cos^2x = -cosx +son^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 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
cosx = -1/2 or cosx = 1

For the first case, cosx = -1/2, x = 120° or x = 240°.

For the second case, cosx = 1, x = 0°.

Therefore, the solutions for the equation cos^2x = -cosx + sin^2x for 0° < x < 360° are x = 0°, x = 120°, and x = 240°.