Solve : $2\sin^3 (x) = \cos (x)$

Here's how I solved it by hand: Notice that if we square both sides, we get $$4 \sin^6 x = \cos^2 x$$ Using the Pythagorean identity, $$4 \sin^6 x + \sin^2 x - 1 = 0$$ Let $y = \sin^2 x$. Our equation becomes $$4y^3 + y - 1 = 0$$ Normally, we would have to use the cubic equation formula, but according to the Rational Root Theorem, we can brute force our way through the eligible rational roots: $\pm 1$,$\pm \frac{1}{2}$, and $\pm \frac{1}{4}$. We notice that $\frac{1}{2}$ is indeed a solution. Dividing out, we see that the remaining roots are roots of $2y^2 + y + 1$, which are complex. Thus, all we have to do is find the numbers for which $$\sin^2 x = \frac{1}{2}$$ You should be able to get that $\pm \frac{\pi}{4}$ and $\pm \frac{5\pi}{4}$ are solutions. We see $-\frac{\pi}{4}$ and $-\frac{5\pi}{4}$ are extraneous solutions. Thus, our solution set is all $\frac{\pi}{4} + k\pi, k\in \mathbb{Z}$

Avoid squaring as it invites Extraneous Roots

$$2=\dfrac{\cos x}{\sin x}\cdot\dfrac1{\sin^2x}=\cot x(\cot^2x+1)$$

$$\cot^3x+\cot x-2=0$$

By observation, $\cot x=1$ is a solution

What about the other two roots?

Generalization:

$$a=\cos^nx\sin^{-2m-n}=\cot^nx(1+\cot^2x)^m$$ where $m$ is an integer

Here $n=1,m=1$