Đáp án:
1) $x =\dfrac{\pi}{6} + k2\pi\quad (k\in\Bbb Z)$
2) $\left[\begin{array}{l}x=-\dfrac{\pi}{6} + k2\pi\\x =\dfrac{\pi}{2} +k2\pi\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
1) $\sin x +\sqrt3\cos x =2$
$\to \dfrac12\sin x +\dfrac{\sqrt3}{2}\cos x = 1$
$\to \sin\left(x +\dfrac{\pi}{3}\right) = 1$
$\to x +\dfrac{\pi}{3} = \dfrac{\pi}{2} + k2\pi$
$\to x =\dfrac{\pi}{6} + k2\pi\quad (k\in\Bbb Z)$
2) $\sin x +\sqrt3\cos x =1$
$\to \dfrac12\sin x +\dfrac{\sqrt3}{2}\cos x = \dfrac12$
$\to \sin\left(x +\dfrac{\pi}{3}\right) = \sin\dfrac{\pi}{6}$
$\to \left[\begin{array}{l}x +\dfrac{\pi}{3} =\dfrac{\pi}{6} + k2\pi\\x +\dfrac{\pi}{3} = \dfrac{5\pi}{6} + k2\pi\end{array}\right.$
$\to \left[\begin{array}{l}x=-\dfrac{\pi}{6} + k2\pi\\x =\dfrac{\pi}{2} +k2\pi\end{array}\right.\quad (k\in\Bbb Z)$
