Question

Answer 1

\(\Leftrightarrow2sin\dfrac{3x}{2}sin\dfrac{x}{2}=2sin\dfrac{3x}{2}cos\dfrac{3x}{2}\)

\(\Leftrightarrow2sin\dfrac{3x}{2}\left(sin\dfrac{x}{2}-cos\dfrac{3x}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\dfrac{3x}{2}=0\\cos\dfrac{3x}{2}=sin\dfrac{x}{2}=cos\left(\dfrac{\pi}{2}-x\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3x}{2}=k\pi\\\dfrac{3x}{2}=\dfrac{\pi}{2}-x+k2\pi\\\dfrac{3x}{2}=x-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{5}+\dfrac{k4\pi}{5}\\x=-\pi+k4\pi\end{matrix}\right.\)