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

$$5\cdot (\frac { 1 }{ 25 } )^{ sin^{ 2 }x }+4\cdot { \left( 5 \right)  }^{ cos2x }=\left( 25 \right) ^{ \frac { sin2x }{ 2 }  }\\ { 5 }^{ -2\sin ^{ 2 }{ x+1 }  }+4\cdot { 5 }^{ 1-2\sin ^{ 2 }{ x }  }={ 5^{ sin2x } }\\ 5\cdot { 5 }^{ 1-2\sin ^{ 2 }{ x }  }={ 5^{ sin2x } }\\ { 5 }^{ 2-2\sin ^{ 2 }{ x }  }=5^{ sin2x }\\ 2-2\sin ^{ 2 }{ x } =\sin { 2x } \\ 1-\sin ^{ 2 }{ x } =\sin { x } \cos { x } \\ \cos ^{ 2 }{ x } =\sin { x } \cos { x } \\ \cos { x } \left( \cos { x } -\sin { x }  \right) =0\\ \cos { x } =0,\cos { x } -\sin { x } =0 $$
Can you proceed from here?
