Giải phương trình: a, 3tan2x -√3 = 0 b, √2sinx + 1= 0 c, -2cos3x + 1 = 0 d, cot2x + 1 = 0

a) $3\tan2x -\sqrt3= 0$ $(*)$

$ĐKXĐ: \, \cos2x \ne  0\Leftrightarrow x \ne \dfrac{\pi}{4} + n\dfrac{\pi}{2}\quad (n\in\Bbb Z)$

$(*)\Leftrightarrow \tan2x = \dfrac{\sqrt3}{3}$

$\Leftrightarrow 2x = \dfrac{\pi}{6} + k\pi$

$\Leftrightarrow x = \dfrac{\pi}{12} + k\dfrac{\pi}{2}\quad (k\in \Bbb Z)$

b) $\sqrt2\sin x + 1 = 0$

$\Leftrightarrow \sin x = -\dfrac{1}{\sqrt2}$

$\Leftrightarrow \left[\begin{array}{l}x = -\dfrac{\pi}{4} + k2\pi\\x = \dfrac{5\pi}{4} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$

c) $-2\cos3x + 1 = 0$

$\Leftrightarrow \cos3x = \dfrac{1}{2}$

$\Leftrightarrow 3x = \pm \dfrac{\pi}{3} + k2\pi$

$\Leftrightarrow x = \pm \dfrac{\pi}{9} + k\dfrac{2\pi}{3}\quad (k\in \Bbb Z)$

d) $\cot2x- 1 = 0 $ $(**)$

$ĐKXĐ: \, \sin2x \ne 0 \Leftrightarrow x \ne n\dfrac{\pi}{2}\quad (n\in \Bbb Z)$

$(**)\Leftrightarrow \cot2x = -1$

$\Leftrightarrow 2x = -\dfrac{\pi}{4} + \pi$

$\Leftrightarrow x = -\dfrac{\pi}{8} + k\dfrac{\pi}{2} \quad (k\in\Bbb Z)$