# Find the solution set of the equation 5. ( 1 25 </mfrac> ) <mr

Find the solution set of the equation $5.\left(\frac{1}{25}{\right)}^{{\mathrm{sin}}^{2}x}+{4.5}^{\mathrm{cos}2x}={25}^{\frac{\mathrm{sin}2x}{2}}$ where $x\in \left[0,2\pi \right]$
My approach :
$5.\left(\frac{1}{25}{\right)}^{{\mathrm{sin}}^{2}x}+{4.5}^{1-2{\mathrm{sin}}^{2}x}={25}^{\frac{\mathrm{sin}2x}{2}}$
Unable to understand how to use $\mathrm{sin}2x$ in R.H.S. to solve further , can you please guide further thanks
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Johnathan Morse
$5\cdot \left(\frac{1}{25}{\right)}^{si{n}^{2}x}+4\cdot {\left(5\right)}^{cos2x}={\left(25\right)}^{\frac{sin2x}{2}}\phantom{\rule{0ex}{0ex}}{5}^{-2{\mathrm{sin}}^{2}x+1}+4\cdot {5}^{1-2{\mathrm{sin}}^{2}x}={5}^{sin2x}\phantom{\rule{0ex}{0ex}}5\cdot {5}^{1-2{\mathrm{sin}}^{2}x}={5}^{sin2x}\phantom{\rule{0ex}{0ex}}{5}^{2-2{\mathrm{sin}}^{2}x}={5}^{sin2x}\phantom{\rule{0ex}{0ex}}2-2{\mathrm{sin}}^{2}x=\mathrm{sin}2x\phantom{\rule{0ex}{0ex}}1-{\mathrm{sin}}^{2}x=\mathrm{sin}x\mathrm{cos}x\phantom{\rule{0ex}{0ex}}{\mathrm{cos}}^{2}x=\mathrm{sin}x\mathrm{cos}x\phantom{\rule{0ex}{0ex}}\mathrm{cos}x\left(\mathrm{cos}x-\mathrm{sin}x\right)=0\phantom{\rule{0ex}{0ex}}\mathrm{cos}x=0,\mathrm{cos}x-\mathrm{sin}x=0$

Patatiniuh
Hint:
Observe
$\begin{array}{rl}{5}^{1-2{\mathrm{sin}}^{2}x}+4\cdot {5}^{1-2{\mathrm{sin}}^{2}x}& ={5}^{2\mathrm{sin}x\mathrm{cos}x}\\ 5\cdot {5}^{1-2{\mathrm{sin}}^{2}x}& ={5}^{2\mathrm{sin}x\mathrm{cos}x}\end{array}$