Sinx - Cotπ/6Cosx=2 B, 2cos^2x -3√2cosx+2=0

b)

$2\cos^2x - 3\sqrt{2} \cos x + 2 = 0$

$<->(\cos x - \sqrt{2})(\cos x - \dfrac{\sqrt{2}}{2}) = 0$

Vậy $\cos x = \sqrt{2}$ (loại) hoặc $\cos x = \dfrac{\sqrt{2}}{2}$

Vậy $x = \pm \dfrac{\pi}{4} + 2k\pi$.