# Sin2x+cosx=0 solve the equation on the interval 0,2pi?

May 14, 2018

$x = \frac{\pi}{2} , \frac{7 \pi}{6} , \frac{3 \pi}{2} , \frac{11 \pi}{6}$

#### Explanation:

Recall the identity $\sin 2 x = 2 \sin x \cos x$. Rewrite the equation with this identity:

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

And note that we can factor out $\cos x :$

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

We now solve the equations $\cos x = 0 , 2 \sin x + 1 = 0 :$

$\cos x = 0 \to x = \frac{\pi}{2} , \frac{3 \pi}{2}$ solve this on the interval $\left[0 , 2 \pi\right) .$

$2 \sin x + 1 = 0$

$2 \sin x = - 1$

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

$x = \frac{7 \pi}{6} , \frac{11 \pi}{6}$ solve this on $\left[0 , 2 \pi\right) .$