giải phtrinh sinx.sin2x.sin3x=1/4.sin4x

Đáp án:

$\left[\begin{array}{l}x = k\dfrac{\pi}{2}\\x= \dfrac{\pi}{8} + k\dfrac{\pi}{4}\end{array}\right.\quad (k\in \Bbb Z)$

Giải thích các bước giải:

$\sin x\sin2x\sin3x = \dfrac{1}{4}\sin4x$

$\Leftrightarrow -\dfrac{1}{2}(\cos4x -\cos2x).\sin2x = \dfrac{1}{2}\sin2x.\cos2x$

$\Leftrightarrow \sin2x(\cos2x + \cos4x -\cos2x)$

$\Leftrightarrow \sin2x.\cos4x = 0$

$\Leftrightarrow \left[\begin{array}{l}\sin2x = 0\\\cos4x= 0\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}x = k\dfrac{\pi}{2}\\x= \dfrac{\pi}{8} + k\dfrac{\pi}{4}\end{array}\right.\quad (k\in \Bbb Z)$