Question

sin 2x + sin x = 0

Answer 1

\(\bf sin^2(x) + sin (x) = 0\implies sin(x)[sin(x)+1]=0\\ \quad \\
\textit{solve for }sin(x)\textit{ and then take }sin^{-1}\)

Answer 2

Is the problem sin (2x) or sin^2 x?

Answer 3

If the problem is actually as you have it written, you could work in this way:\[\sin (2x) +\sin x=0 \implies 2\cos x \sin x + \sin x = 0\]\[\implies \cos x = - \frac{1}{2} \implies x=\pm \cos^{-1}(-\frac{1}{2}) + 2k \pi~~~~k \in \mathbb{Z}\]From this we get\[x= \pm \frac{2\pi}{3} + 2k \pi \]

Answer 4

Also note the solution \[x=2k \pi\]also solves the problem.