Some help on trigonometric equation So I have $\sin^3x = \frac 34 \sin x$.
Can you expand so the answer is either $\sin x(\sin^2x +\frac 34)$ which leads to
 the answer $\frac 12 + 2n\pi$ or that $\sin^3x = \frac 14(3\sin x-\sin^3x) - \frac 34\sin x$ which leads to the answer $0 + 2n \pi$.
Is that correct by any chance?

Notice, we have $$\sin^3x = \frac 34 \sin x$$ $$\sin^3x -\frac 34 \sin x=0$$ $$\sin x\left(\sin^2 x-\frac{3}{4}\right)=0$$ $$\text{if}\ \sin x=0\ \implies \color{blue}{x=n\pi}$$ $$\text{if}\ \sin^2 x-\frac{3}{4}\iff \sin^2x=\left(\frac{\sqrt{3}}{2}\right)^2 \iff \sin^2x=\left(\sin \frac{\pi}{3}\right)^2\ \implies \color{blue}{x=n\pi\pm \frac{\pi}{3}}$$ Where, $n$ is an integer. 
