# How do you solve sin2x-cosx=0  over the interval 0 to 2pi?

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mason m Share
May 9, 2016

$x = \frac{\pi}{6} , \frac{\pi}{2} , \frac{5 \pi}{6} , \frac{3 \pi}{2}$

#### Explanation:

Use the identity $\sin 2 x = 2 \sin x \cos x$.

$2 \sin x \cos x - \cos x = 0$

Factor a $\cos x$ term on the left hand side.

$\cos x \left(2 \sin x - 1\right) = 0$

Set both of these terms equal to $0$.

$\cos x = 0 \text{ "=>" } x = \frac{\pi}{2} , \frac{3 \pi}{2}$

$2 \sin x - 1 = 0 \text{ "=>" "sinx=1/2" "=>" } x = \frac{\pi}{6} , \frac{5 \pi}{6}$

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