Đáp án:
\(S = \left\{-\dfrac{\pi}{6} +k2\pi;\ \dfrac{7\pi}{6} + k2\pi;\ \dfrac{\pi}{4} + k\dfrac{\pi}{2}\ \Bigg|\ k\in\Bbb Z\right\}\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad \sin3x +\cos2x - \sin x = 0\\
\Leftrightarrow (3\sin x - 4\sin^3x) + (1 - 2\sin^2x) - \sin x =0\\
\Leftrightarrow 4\sin^3x +2\sin^2x -2\sin x - 1 =0\\
\Leftrightarrow (2\sin x +1)(2\sin^2x - 1) =0\\
\Leftrightarrow (2\sin x + 1).\cos2x = 0\\
\Leftrightarrow \left[\begin{array}{l}\sin x = -\dfrac12\\\cos2x = 0\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = -\dfrac{\pi}{6} +k2\pi\\x = \dfrac{7\pi}{6} + k2\pi\\x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy}\ S = \left\{-\dfrac{\pi}{6} +k2\pi;\ \dfrac{7\pi}{6} + k2\pi;\ \dfrac{\pi}{4} + k\dfrac{\pi}{2}\ \Bigg|\ k\in\Bbb Z\right\}
\end{array}\)
