Đáp án:
\[\left[ \begin{array}{l}
x = \dfrac{\pi }{2} = k\pi \\
x = \dfrac{{k2\pi }}{3}\\
x = \pi + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\sin x + \sin y = 2\sin \dfrac{{x + y}}{2}.\cos \dfrac{{x - y}}{2}\\
\sin 2x = 2\sin x.\cos x\\
\sin x + \sin 2x + \sin 3x = 0\\
\Leftrightarrow \left( {\sin 3x + \sin x} \right) + \sin 2x = 0\\
\Leftrightarrow 2\sin \dfrac{{3x + x}}{2}.\cos \dfrac{{3x - x}}{2} + 2\sin x.\cos x = 0\\
\Leftrightarrow 2\sin 2x.\cos x + 2\sin x.\cos x = 0\\
\Leftrightarrow 2\cos x.\left( {\sin 2x + \sin x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\sin 2x + \sin x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\sin 2x = - \sin x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\sin 2x = \sin \left( { - x} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
2x = - x + k2\pi \\
2x = \pi - \left( { - x} \right) + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
3x = k2\pi \\
x = \pi + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} = k\pi \\
x = \dfrac{{k2\pi }}{3}\\
x = \pi + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
