Đáp án:
Giải thích các bước giải:
$\quad 3\cos2x + 10\sin x + 1 = 0$
$\Leftrightarrow 3(1-2\sin^2x) + 10\sin x + 1 = 0$
$\Leftrightarrow 3\sin^2x - 5\sin x - 2 = 0$
$\Leftrightarrow (3\sin x +1)(\sin x -2)= 0$
$\Leftrightarrow \left[\begin{array}{l}\sin x = -\dfrac13\\\sin x = 2\quad (vn)\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \arcsin\left(-\dfrac13\right) + k2\pi\\x = \pi - \arcsin\left(-\dfrac13\right) + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$
Ta lại có: $-\dfrac{\pi}{2}< x < \dfrac{\pi}{2}$
$\Leftrightarrow \left[\begin{array}{l}-\dfrac{\pi}{2}< \arcsin\left(-\dfrac13\right) + k2\pi< \dfrac{\pi}{2}\\-\dfrac{\pi}{2}< \pi - \arcsin\left(-\dfrac13\right) + k2\pi < \dfrac{\pi}{2}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}-\dfrac{1}{4} - \dfrac{1}{2\pi}\arcsin\left(-\dfrac13\right)< k< \dfrac{1}{4} - \dfrac{1}{2\pi}\arcsin\left(-\dfrac13\right)\\-\dfrac{3}{4} + \dfrac{1}{2\pi}\arcsin\left(-\dfrac13\right)< k< -\dfrac{1}{4}+\dfrac{1}{2\pi}\arcsin\left(-\dfrac13\right)\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}k = 0\\k \in\varnothing\end{array}\right.$
Vậy $x = \arcsin\left(-\dfrac13\right)$
