Find the solution set of the equation $5.(\frac{1}{25})^{\sin^2x}+4.5^{\cos2x}=25^{\frac{\sin2x}{2}}$ Problem : 
Find the solution set of the equation $5.(\frac{1}{25})^{\sin^2x}+4.5^{\cos2x}=25^{\frac{\sin2x}{2}}$ where $x \in [0,2\pi]$
My approach : 
$5.(\frac{1}{25})^{\sin^2x}+4.5^{1-2\sin^2x}=25^{\frac{\sin2x}{2}}$
Unable to understand how to use $\sin2x$ in R.H.S. to solve further , can you please guide further thanks

Hint:
Observe
\begin{align*}
5^{1-2\sin^2 x}+4\cdot 5^{1-2\sin^2 x}&=5^{2\sin x\cos x}\\
5\cdot 5^{1-2\sin^2 x}&=5^{2\sin x\cos x}\\
\end{align*}
