\(sin\left(\dfrac{3\pi}{5}+2x\right)=2sin^2\left(\dfrac{\pi}{5}-x\right)\)
\(\Leftrightarrow sin\left(\dfrac{3\pi}{5}+2x\right)=1-cos\left(\dfrac{2\pi}{5}-2x\right)\)
\(\Leftrightarrow sin\left(\dfrac{3\pi}{5}+2x\right)=1-sin\left(\dfrac{2\pi}{5}-2x+\dfrac{\pi}{2}\right)\)
\(\Leftrightarrow sin\left(\dfrac{3\pi}{5}+2x\right)+sin\left(\dfrac{9\pi}{10}-2x\right)=1\)
\(\Leftrightarrow2sin\left(\dfrac{\dfrac{3\pi}{5}+\dfrac{9\pi}{10}}{2}\right)cos\left(2x+\dfrac{\dfrac{3\pi}{5}-\dfrac{9\pi}{10}}{2}\right)=1\)
\(\Leftrightarrow2sin\left(\dfrac{3\pi}{4}\right)cos\left(2x-\dfrac{3\pi}{20}\right)=1\)
\(\Leftrightarrow\sqrt{2}cos\left(2x-\dfrac{3\pi}{20}\right)=1\)
\(\Leftrightarrow cos\left(2x-\dfrac{3\pi}{20}\right)=\dfrac{1}{\sqrt{2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{3\pi}{20}=\dfrac{\pi}{4}+k2\pi\\2x-\dfrac{3\pi}{20}=\dfrac{-\pi}{4}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{5}+k\pi\\x=\dfrac{-\pi}{20}+k\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
