How do you solve -sin^2 x=2cosx-2

Question

anonymous · Answer

$$-2\sin^2x=2cosx-2$$

zepdrix · Answer

Adding 2sin^2x to each side gives us,$$\Large 0=2\sin^2x+2\cos x-2$$Divide both sides by 2,$$\Large 0=\sin^2x+\cos x-1$$

From here we want to remember our `Square Identity`:$$\large \sin^2x+\cos^2x=1\qquad\to\qquad \color{royalblue}{\sin^2x=1-\cos^2x}$$Applying this identity to our problem,$$\Large 0=\color{royalblue}{\sin^2x}+\cos x-1$$Gives us,$$\Large 0=\color{royalblue}{1-\cos^2x}+\cos x-1$$

zepdrix · Answer

Just requires a few more steps from here.
Confused by any of that?

anonymous · Answer

no, but what do i do after getting$$0=cosx-\cos^2x$$

anonymous · Answer

o, is x=pi/2

zepdrix · Answer

Factor cosx out of each term.$$\Large 0=\cos x(1-\cos x)$$Then by the Zero Factor Property we set each term equal to zero and solve separately.$$\Large 0=\cos x \qquad\qquad\qquad 0=1-\cos x$$

x=pi/2 sounds correct, that is ONE of our solutions :)