1 câu trả lời
Đáp án:$\left[ \begin{array}{l}
x = \frac{{k\pi }}{2}\\
x = \frac{{2\pi }}{3} + k2\pi \\
x = - \frac{{2\pi }}{3} + k2\pi
\end{array} \right.\left( {k \in Z} \right)$
Giải thích các bước giải:
$\begin{array}{l}
\sin x + \sin 2x + \sin 3x = 0\\
\Rightarrow \sin x + \sin 3x + \sin 2x = 0\\
\Rightarrow 2\sin 2x.\cos \left( { - x} \right) + \sin 2x = 0\\
\Rightarrow \sin 2x\left( {2\cos x + 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\sin 2x = 0\\
\cos x = \frac{{ - 1}}{2}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
2x = k\pi \\
x = \frac{{2\pi }}{3} + k2\pi \\
x = - \frac{{2\pi }}{3} + k2\pi
\end{array} \right.\left( {k \in Z} \right) \Rightarrow \left[ \begin{array}{l}
x = \frac{{k\pi }}{2}\\
x = \frac{{2\pi }}{3} + k2\pi \\
x = - \frac{{2\pi }}{3} + k2\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}$