Find the solution set of the equation $5.(\frac{1}{25}{)}^{{\mathrm{sin}}^{2}x}+{4.5}^{\mathrm{cos}2x}={25}^{\frac{\mathrm{sin}2x}{2}}$ where $x\in [0,2\pi ]$

My approach :

$5.(\frac{1}{25}{)}^{{\mathrm{sin}}^{2}x}+{4.5}^{1-2{\mathrm{sin}}^{2}x}={25}^{\frac{\mathrm{sin}2x}{2}}$

Unable to understand how to use $\mathrm{sin}2x$ in R.H.S. to solve further , can you please guide further thanks

My approach :

$5.(\frac{1}{25}{)}^{{\mathrm{sin}}^{2}x}+{4.5}^{1-2{\mathrm{sin}}^{2}x}={25}^{\frac{\mathrm{sin}2x}{2}}$

Unable to understand how to use $\mathrm{sin}2x$ in R.H.S. to solve further , can you please guide further thanks