Đáp án:
\(S = \left\{- \dfrac{\pi}{6} + k2\pi;\ k2\pi;\ \dfrac{7\pi}{6} + k2\pi\ \Bigg|\ k\in\Bbb Z\right\}\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad \sin2x - 2\sin x + \cos x - 1 =0\\
\Leftrightarrow 2\sin x\cos x - 2\sin x+ \cos x - 1 =0\\
\Leftrightarrow 2\sin x(\cos x - 1) + \cos x - 1 =0\\
\Leftrightarrow (\cos x - 1)(2\sin x + 1) = 0\\
\Leftrightarrow \left[\begin{array}{l}\cos x = 1\\\sin x = - \dfrac12\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = k2\pi\\x = - \dfrac{\pi}{6} + k2\pi\\x = \dfrac{7\pi}{6} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy}\ S = \left\{- \dfrac{\pi}{6} + k2\pi;\ k2\pi;\ \dfrac{7\pi}{6} + k2\pi\ \Bigg|\ k\in\Bbb Z\right\}
\end{array}\)
