Đáp án:
$\left[\begin{array}{l}x = \dfrac{\pi}{4} - \arccos\dfrac{3}{\sqrt{10}} + k\pi\\x = \dfrac{\pi}{4} +k\pi\end{array}\right. \quad (k \in \Bbb Z)$
Giải thích các bước giải:
$\sin^2x + 3\sin x\cos x + 2\cos^2x = 0$
$\Leftrightarrow \dfrac{3}{2}\sin2x + \cos^2x + 1 = 0$
$\Leftrightarrow \dfrac{3}{2}\sin2x + \dfrac{1 + \cos2x}{2} + 1 = 0$
$\Leftrightarrow 3\sin2x +\cos2x = 3$
$\Leftrightarrow \dfrac{3}{\sqrt{10}}\sin2x + \dfrac{1}{\sqrt{10}}\cos2x = \dfrac{3}{\sqrt{10}}$
Ta có:
$\left(\dfrac{3}{\sqrt{10}}\right)^2 + \left(\dfrac{1}{\sqrt{10}}\right)^2 = 1$
$\Rightarrow Đặt \,\,\begin{cases}\cos\alpha = \dfrac{3}{\sqrt{10}}\\\sin\alpha = \dfrac{1}{\sqrt{10}}\end{cases}\Rightarrow \alpha = \arccos\dfrac{3}{\sqrt{10}}$
Ta được:
$\dfrac{3}{\sqrt{10}}\sin2x + \dfrac{1}{\sqrt{10}}\cos2x = \dfrac{3}{\sqrt{10}}$
$\Leftrightarrow \sin2x\cos\alpha + \cos2x\sin\alpha = \cos\alpha$
$\Leftrightarrow \sin(2x + \alpha) = \sin\left(\dfrac{\pi}{2} - \alpha\right)$
$\Leftrightarrow \left[\begin{array}{l}2x + \alpha = \dfrac{\pi}{2} - \alpha + k2\pi\\2x + \alpha = \dfrac{\pi}{2} + \alpha +k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{4} - \alpha + k\pi\\x = \dfrac{\pi}{4} +k\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{4} - \arccos\dfrac{3}{\sqrt{10}} + k\pi\\x = \dfrac{\pi}{4} +k\pi\end{array}\right. \quad (k \in \Bbb Z)$
