A trigonometric equation is given such that $$\sqrt{1 + \frac{3\cos2x}{{\sqrt {\sin^4x+4\cos^2x}} + \sqrt{\cos^4x+4\sin^2x}}}=\sqrt2 \sin^{-1}(\sin2x)$$ $x \in [-\pi,\pi]$
It is to be proven that one of the roots of this equation lies in the interval $(\frac{\pi}{8}, \frac{\pi}{6})$.
I took the subset $[-\frac{\pi}{4},\frac{\pi}{4}]$ so that $sin^{-1}(sin2x)=2x$ holds true. Now I squared both sides to obtain $$ \frac{3\cos2x}{{\sqrt {\sin^4x+4\cos^2x}} + \sqrt{\cos^4x+4\sin^2x}}= 8x^2-1 $$
With another small manipulation, $$ {\sqrt {\sin^4x+4\cos^2x}} - \sqrt{\cos^4x+4\sin^2x}= 8x^2-1 $$
How should I proceed after this? Also what is the total nuber of solutions to the above equation?