# How do you solve the equation on the interval [0,2pi) for sin 3x = -sinx?

Aug 24, 2016

The Soln. Set is $\left\{0 , \frac{\pi}{2} , \pi , 3 \frac{\pi}{2}\right\}$.

#### Explanation:

We recall that, $\sin 3 x = 3 \sin x - 4 {\sin}^{3} x$, so, the given eqn. becomes,

$3 \sin x - 4 {\sin}^{3} x + \sin x = 0$.

$\therefore 4 \sin x - 4 {\sin}^{3} x = 0$.

$\therefore 4 \sin x \left(1 - {\sin}^{2} x\right) = 0$.

$\therefore 4 \sin x {\cos}^{2} x = 0$.

$\therefore \sin x = 0 , \cos x = 0$.

$\therefore x = k \pi , \mathmr{and} , x = \left(2 k + 1\right) \frac{\pi}{2} , k \in \mathbb{Z}$.

Hence, the Soln. Set is $\left\{k \pi\right\} \cup \left\{\left(2 k + 1\right) \frac{\pi}{2}\right\} , k \in \mathbb{Z}$.

:. "But, x" in [0,2pi), &, sinx=0 rArr x=0,pi, while,

for $\cos x = 0 , \text{such x are} \frac{\pi}{2} , 3 \frac{\pi}{2}$.