Đáp án:

\(\left[ \begin{array}{l} x=\dfrac{\pi}{2}+k.2\pi\\\left[ \begin{array}{l}x=\dfrac{\pi}{6}+k.2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{array} \right.\end{array} \right.\) \((k \epsilon Z)\)

Giải thích các bước giải:

 \(\cos 2x +3\sin x-2=0\)

\(\Leftrightarrow 1-2\sin^{2} x+3\sin x-2=0\)

\(\Leftrightarrow -2\sin^{2} x +3\sin x-1=0\) \((-1 \leq \sin x \leq 1)\)

\(\Leftrightarrow \) \(\left[ \begin{array}{l}\sin x=1\\\sin x=\dfrac{1}{2}\end{array} \right.\) 

\(\Leftrightarrow \) \(\left[ \begin{array}{l} x=\dfrac{\pi}{2}+k.2\pi\\\sin x=\sin \dfrac{\pi}{6}\end{array} \right.\) 

\(\Leftrightarrow \) \(\left[ \begin{array}{l} x=\dfrac{\pi}{2}+k.2\pi\\\left[ \begin{array}{l}x=\dfrac{\pi}{6}+k.2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{array} \right.\end{array} \right.\) \((k \epsilon Z)\)

Đáp án:

$\left[\begin{array}{l}x = \dfrac{\pi}{2} + k2\pi\\x =\dfrac{\pi}{6}+k2\pi\\x = \dfrac{5\pi}{6}+ k2\pi\end{array}\right.\quad (k\in \Bbb Z)$

Giải thích các bước giải:

$cos2x + 3sinx - 2 = 0$

$\Leftrightarrow 1 - 2sin^2x + 3sinx - 2 = 0$

$\Leftrightarrow 2sin^2x - 3sinx +1 = 0$

$\Leftrightarrow \left[\begin{array}{l}sinx = 1\\sinx =\dfrac{1}{2}\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{2} + k2\pi\\x =\dfrac{\pi}{6}+k2\pi\\x = \dfrac{5\pi}{6}+ k2\pi\end{array}\right.\quad (k\in \Bbb Z)$