giải phtrinh sinx.sin2x.sin3x=1/4.sin4x 10/07/2021 Bởi Katherine 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: $\sin x\sin2x\sin3x = \dfrac{1}{4}\sin4x$ $\Leftrightarrow -\dfrac{1}{2}(\cos4x -\cos2x).\sin2x = \dfrac{1}{2}\sin2x.\cos2x$ $\Leftrightarrow \sin2x(\cos2x + \cos4x -\cos2x)$ $\Leftrightarrow \sin2x.\cos4x = 0$ $\Leftrightarrow \left[\begin{array}{l}\sin2x = 0\\\cos4x= 0\end{array}\right.$ $\Leftrightarrow \left[\begin{array}{l}x = k\dfrac{\pi}{2}\\x= \dfrac{\pi}{8} + k\dfrac{\pi}{4}\end{array}\right.\quad (k\in \Bbb Z)$ Bình luận
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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:
$\sin x\sin2x\sin3x = \dfrac{1}{4}\sin4x$
$\Leftrightarrow -\dfrac{1}{2}(\cos4x -\cos2x).\sin2x = \dfrac{1}{2}\sin2x.\cos2x$
$\Leftrightarrow \sin2x(\cos2x + \cos4x -\cos2x)$
$\Leftrightarrow \sin2x.\cos4x = 0$
$\Leftrightarrow \left[\begin{array}{l}\sin2x = 0\\\cos4x= 0\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = k\dfrac{\pi}{2}\\x= \dfrac{\pi}{8} + k\dfrac{\pi}{4}\end{array}\right.\quad (k\in \Bbb Z)$