Đáp án: $\left[\begin{array}{l}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\\x=\dfrac{-\pi}{2}+k2\pi\end{array}\right.\,\,\,\,\,\left(k\in\mathbb{Z}\right)$
Giải thích:
$\,\,\,\,\,\,\,2{{\cos }^{2}}x-\sin x-1=0$
$\Leftrightarrow 2\left( 1-{{\sin }^{2}}x \right)-\sin x-1=0$
$\Leftrightarrow -2{{\sin }^{2}}x-\sin x+1=0$
$\Leftrightarrow\left[\begin{array}{l}\sin x=\dfrac{1}{2}\\\sin x=-1\end{array}\right.$
$\Leftrightarrow\left[\begin{array}{l}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\\x=\dfrac{-\pi}{2}+k2\pi\end{array}\right.\,\,\,\,\,\left(k\in\mathbb{Z}\right)$
