Đáp án:
$\left\{\begin{array}{I}x= \pm\dfrac{2\pi}3+k2\pi\\ x=\dfrac{\arctan \sqrt 2}2+\dfrac{k\pi}2\end{array}\right.$ $(k\in\mathbb Z)$
Giải thích các bước giải:
$\sin x+\sin 2x+\sin3x=\sqrt2(\cos x+\cos2x+\cos 3x)$
$(\sin x+ \sin 3x)+\sin2x= \sqrt 2.( \cos x+\cos3x+\cos2x)$
$2\sin2x.\cos x+\sin2x= \sqrt 2.( 2\cos2x.\cos x+ \cos2x)$
$\sin2x.( 2\cos x+1)=\sqrt 2. \cos2x. (2\cos x+1)$
$(2\cos x+1).(\sin2x- \sqrt 2. \cos2x)=0$
$\left[\begin{array}{I}\cos x= -\dfrac12\\ \tan2x= \sqrt 2\end{array}\right. \Leftrightarrow\left[\begin{array}{I}x= \pm\dfrac{2\pi}3+k2\pi\\ x=\dfrac{\arctan \sqrt 2}2+\dfrac{k\pi}2\end{array}\right.$ $(k\in\mathbb Z)$
