Đáp án:
b. \(\left[ \begin{array}{l}
x = \dfrac{{k\pi }}{3}\\
x = \dfrac{\pi }{3} + k\pi \\
x = - \dfrac{\pi }{3} + k\pi
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.3\sin x + 2\sin x\cos x = 0\\
\to \sin x\left( {3 + 2\cos x} \right) = 0\\
\to \left[ \begin{array}{l}
\sin x = 0\\
\cos x = - \dfrac{3}{2}\left( l \right)
\end{array} \right.\\
\to x = k\pi \left( {k \in Z} \right)\\
b.\sin x + \sin 3x + \sin 5x = 0\\
\to \sin 3x + 2\sin 3x.\cos 2x = 0\\
\to \sin 3x\left( {1 + 2\cos 2x} \right) = 0\\
\to \left[ \begin{array}{l}
\sin 3x = 0\\
\cos 2x = - \dfrac{1}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
3x = k\pi \\
2x = \dfrac{{2\pi }}{3} + k2\pi \\
2x = - \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{3}\\
x = \dfrac{\pi }{3} + k\pi \\
x = - \dfrac{\pi }{3} + k\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)
