giải phương trình sau: sin7x.sinx=sin5x.sin3x

$\sin7x.\sin x=\sin5x.\sin3x$

$\Leftrightarrow \dfrac{-1}{2}(\cos8x-\cos x)=\dfrac{-1}{2}(\cos 8x-\cos 2x)$

$\Leftrightarrow \cos2x=\cos x$

$\Leftrightarrow 2x=\pm x+k2\pi$

$\Leftrightarrow x=k2\pi$, $x=\dfrac{k2\pi}{3}$

Vậy $x=\dfrac{k2\pi}{3}$