Solve for $x$ in $x\sqrt{1-4x^2}+2x\sqrt{1-x^2}=\frac{\sqrt 3}2$ Find out the value of $x$ in $x\sqrt{1-4x^2}+2x\sqrt{1-x^2}=\dfrac{\sqrt 3}2$.
I tried squaring both the sides but it only makes it more complicated.  Is there any other way?

Put $2x = \cos(\alpha), x = \cos(\beta) \implies \sin(\alpha)\cos(\beta) + \sin(\beta)\cos(\alpha) = \dfrac{\sqrt{3}}{2}\implies \sin(\alpha+\beta) = \dfrac{\sqrt{3}}{2}\implies \alpha+\beta=\dfrac{\pi}{3}, \dfrac{2\pi}{3}\implies \beta = \dfrac{\pi}{3} - \alpha\implies x = \cos(\dfrac{\pi}{3} -\alpha)= \dfrac{1}{2}(2x)+\dfrac{\sqrt{3}}{2}\sin(\alpha)\implies \sin(\alpha) = 0 \implies \cos(\alpha) = 1\implies x = \dfrac{1}{2}$. The other case that $\alpha+\beta = \dfrac{2\pi}{3}$, I leave it for you to finish.
