$\text{Đáp án + Giải thích các bước giải:}$

$a)$

$\cos \left(3x - \dfrac{\pi}{4}\right) = -\dfrac{\sqrt{2}}{2} \quad \Leftrightarrow \cos \left(3x - \dfrac{\pi}{4}\right) = \cos \dfrac{3\pi}{4}$

$\Leftrightarrow \left[\begin{aligned} 3x - \dfrac{\pi}{4} &= \dfrac{3\pi}{4} + k2\pi \\ 3x - \dfrac{\pi}{4} &= -\dfrac{3\pi}{4} + k2\pi \end{aligned}\right.$

$\Leftrightarrow \left[\begin{aligned} 3x &= \pi + k2\pi \\ 3x &= -\dfrac{\pi}{2} + k2\pi \end{aligned}\right.$

$\Leftrightarrow \left[\begin{aligned} x &= \dfrac{\pi}{3} + \dfrac{k2\pi}{3} \\ x &= -\dfrac{\pi}{6} + \dfrac{k2\pi}{3} \end{aligned}\right. \quad (k \in \mathbb{Z})$

$b)$

$2\sin^2 x - 1 + \cos 3x = 0 \quad \Leftrightarrow -\cos 2x + \cos 3x = 0$

$\Leftrightarrow \cos 3x = \cos 2x$

$\Leftrightarrow \left[\begin{aligned} 3x &= 2x + k2\pi \\ 3x &= -2x + k2\pi \end{aligned}\right.$

$\Leftrightarrow \left[\begin{aligned} x &= k2\pi \\ 5x &= k2\pi \end{aligned}\right.$

$\Leftrightarrow \left[\begin{aligned} x &= k2\pi \\ x &= \dfrac{k2\pi}{5} \end{aligned}\right. \quad (k \in \mathbb{Z})$

$c)$

ĐKXĐ:

$2x + \dfrac{\pi}{5} \neq \dfrac{\pi}{2} + k\pi \Rightarrow x \neq \dfrac{3\pi}{20} + \dfrac{k\pi}{2}$

$x - \dfrac{\pi}{6} \neq \dfrac{\pi}{2} + k\pi \Rightarrow x \neq \dfrac{2\pi}{3} + k\pi$

$\tan \left(2x + \dfrac{\pi}{5}\right) = \tan \left(x - \dfrac{\pi}{6}\right) \quad \Leftrightarrow 2x + \dfrac{\pi}{5} = x - \dfrac{\pi}{6} + k\pi$

$\Leftrightarrow x = -\dfrac{11\pi}{30} + k\pi \quad (k \in \mathbb{Z})$

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