# How do you solve sin2x-cosx=0?

Nov 25, 2016

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

#### Explanation:

Before we solve, we need to note an identity:

$\sin 2 x = 2 \sin x \cos x$

Using the identity from above, rewrite the equation.

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

Now factor out a $\cos x$.

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

Now take each factor and set it equal to zero.

$\cos x = 0$

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



$2 \sin x - 1 = 0$

$2 \sin x = 1$

$\sin x = \frac{1}{2}$

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