a) Ta có
$2sin^2x - 5\sin x + 2 = 0$
Ptrinh có nghiệm $\sin x = 2$ (loại) hoặc $\sin x = \dfrac{1}{2}$
Vậy $x = \dfrac{\pi}{6} + 2k\pi$ hoặc $x = \dfrac{5\pi}{6} + 2k\pi$.
b) Chia 2 vế cho 2 ta có
$\dfrac{1}{2} \sin(2x) + \dfrac{\sqrt{3}}{2} \cos(2x) = \dfrac{1}{2}$
$<-> \cos(2x) \cos(\dfrac{\pi}{6}) + \sin(\dfrac{\pi}{6}) \sin(2x) = \cos(\dfrac{\pi}{3})$
$<-> \cos(2x - \dfrac{\pi}{6}) = \cos(\dfrac{\pi}{3})$
Vậy $2x - \dfrac{\pi}{6} = \pm \dfrac{\pi}{3} + 2k\pi$.
Do đó $x = \dfrac{\pi}{4} + k\pi$ hoặc $x = -\dfrac{\pi}{12} + k\pi$.
