Đáp án:
$\left\{\begin{array}{I}x =k\dfrac{\pi}3\\x = \dfrac{\pi}3+k\pi\\x =-\dfrac{\pi}3+k\pi\end{array}\right.$ $(k\in\mathbb Z)$
Lời giải:
$\sin5x + \sin x = 2\sin\dfrac{5x + x}2\cos\dfrac{5x - x}2$
$= 2\sin3x\cos2x$.
Phương trình:
$\sin x + \sin 3x + \sin 5x = 0$
$\Leftrightarrow 2\sin3x\cos2x + \sin3x = 0$
$\Leftrightarrow \sin3x(2\cos2x + 1) = 0$
$\Leftrightarrow \left[\begin{array}{I}\sin3x = 0\\\cos2x = -\dfrac12\end{array}\right.\Leftrightarrow \left[\begin{array}{I}3x =k\pi\\2x = \dfrac{2\pi}3+k2\pi\\2x =-\dfrac{2\pi}3+k2\pi\end{array}\right.$ $(k\in\mathbb Z)$
Vậy phương trình có nghiệm
$\left\{\begin{array}{I}x =k\dfrac{\pi}3\\x = \dfrac{\pi}3+k\pi\\x =-\dfrac{\pi}3+k\pi\end{array}\right.$ $(k\in\mathbb Z)$.
