Solve the equation $\frac{\sqrt {3}}{2}\sin(x) -\cos x=\cos^2x$ Solve the equation $\frac{\sqrt {3}}{2}\sin x-\cos x=\cos^2x$
My approach $\cos^2x=1-\sin^2x $
$\frac{\sqrt {3}}{2}\sin x-\cos^2x =\cos x$
$\frac{\sqrt {3}}{2}\sin x+\sin^2x -1=\cos x$
$\sqrt {3}\sin x+2\sin^2x-2=2\cos x$
Though the equation comes in form of $\sin x$ from here onward after squaring still not getting the answer.

A straightforward approach is indeed to square the equation: 
$$\frac{\sqrt{3}}{2}\sin x = \cos^2 x +\cos x$$
and replace $\sin^2 x$ by $1-\cos^2 x$. You obtain a quartic equation in $\cos x$. Maybe not the most elegant solution, but you cannot go wrong with it. Also, one root is clear without calculation: $\cos x = 1/2$ is fine. So you can surely reduce to a cubic equation. (And possibly even further to a quadratic one.)
