# How do you solve the identity sin(x) + sin (2x) = 0?

Hence $\sin 2 x = 2 \sin x \cdot \cos x$ we have that

sinx+sin2x=0=>sinx+2sinxcosx=0=>sinx(1+2cosx)=0=> sinx=0 or cosx=-1/2

From $\sin x = 0 \implies x = k \cdot \pi$ , $k \setminus \in N$ and

from $\cos x = - \frac{1}{2} \implies \cos x = \cos \left(2 \frac{\pi}{3}\right) \implies x = 2 \cdot k \cdot \pi \pm \frac{2 \cdot \pi}{3}$

Oct 14, 2015

$x = k \pi \vee x = \frac{2 \pi}{3} + 2 m \pi \vee x = \frac{4 \pi}{3} + 2 n \pi$
$k , m , n \in Z$

#### Explanation:

First of all, this is not identity, it's an equation.

Using trigonometric identity:
$\sin 2 x = 2 \sin x \cos x$

our equation becomes:

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

$\sin x = 0 \vee 1 + 2 \cos x = 0$

$\sin x = 0 \vee \cos x = - \frac{1}{2}$

$x = k \pi \vee x = \frac{2 \pi}{3} + 2 m \pi \vee x = \frac{4 \pi}{3} + 2 n \pi$
$k , m , n \in Z$