# Solving a trig equation

How would I solve the following trig equation

$\sin^2x=1-\cos(x)$

I have to write the solution in radians.

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Recall $$\sin^2x = 1 - \cos^2 x$$

So you can write the equation as a quadratic, in $\cos x$:

$$\sin^2x=1-\cos(x)\iff 1 - \cos^2x = 1 - \cos x$$ $$\iff \cos^2 x - \cos x = 0 \iff \cos x(\cos x - 1) = 0$$

Let $u = \cos x$, e.g. $$u(u-1) = 0 \implies u = 0 \text{ or}\;u = 1$$

$$u = \cos x = 0 \; \implies\; x = \frac{(2k + 1)\pi}{2},\; k \in \mathbb Z$$ $$u = \cos x = 1 \; \implies\; x = 2k\pi, \; k \in \mathbb Z$$

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ok I will try that. –  Fernando Martinez Feb 19 '13 at 1:29
I got $cos^2(x)-cos(x)=0$ would that mean that x=pi and x=pi/2 –  Fernando Martinez Feb 19 '13 at 1:32
Oh I see your post never mind, thanks. –  Fernando Martinez Feb 19 '13 at 1:33
Shouldn't it be $x=2k\pi$? $\cos\pi=-1$ –  Mike Feb 19 '13 at 5:42
Yes, @Mike, thanks for catching the typo! –  amWhy Feb 19 '13 at 12:55