Solutions to $\sqrt{3 - \tan^2(\frac{3x}{2})} \cdot \sin x - \cos x = 2$. $$\sqrt{3 - \tan^2(\frac{3x}{2})} \cdot \sin x - \cos x = 2.$$
How can such an equation be solved? I don't know where to start?
My attempt: $$(3 - \frac{\sin^2 3 y}{\cos^2 3 y}) \cdot 2 \cos y \sin y = (3 - 2 \sin^2 y)^2$$, where $y = \frac{x}{2}$. This does not simplify anything at all, I do not understand what to do next

By transferring $\cos x$ to the RHS and squaring, we have
$$(3 - \tan^2\tfrac32x)\sin ^2x=(2+\cos x)^2.$$
Using the fact that $\cos^2x+\sin^2x=1$, some rearrangement gives
$$(\tan\tfrac32x\sin x)^2+(1+2\cos x)^2=0.$$
A real solution to this equation would simultaneously require, modulo $2\pi$, that $x=0$ and $x=\pm\frac23\pi$. Since these requirements are inconsistent, the original equation has no solution.
