Đáp án:
$\alpha.\beta = - \dfrac{\pi^2}{12}$
Giải thích các bước giải:
$\begin{array}{l}\sin x + \sqrt3\cos x = 1\\ \Leftrightarrow \dfrac{1}{2}\sin x + \dfrac{\sqrt3}{2}\cos x = \dfrac{1}{2}\\ \Leftrightarrow \cos\left(x - \dfrac{\pi}{6}\right) = \cos\dfrac{\pi}{3}\\ \Leftrightarrow \left[\begin{array}{l}x - \dfrac{\pi}{6} = \dfrac{\pi}{3} +k2\pi\\x - \dfrac{\pi}{6} = -\dfrac{\pi}{3} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{2} +k2\pi\\x =- \dfrac{\pi}{6} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)\\ \Rightarrow \begin{cases}\alpha = -\dfrac{\pi}{6}\\ \beta = \dfrac{\pi}{2}\end{cases}\Rightarrow \alpha.\beta = - \dfrac{\pi^2}{12} \end{array}$
