$\begin{array}{l} b)4{\cos ^2}x - 2\left( {\sqrt 3 - \sqrt 2 } \right)\cos x = \sqrt 6 \\ \Leftrightarrow 4{\cos ^2}x - 2\left( {\sqrt 3 - \sqrt 2 } \right)\cos x - \sqrt 6 = 0\\ \Delta ' = {\left( {\sqrt 3 - \sqrt 2 } \right)^2} - 4\left( { - \sqrt 6 } \right) = 5 - 2\sqrt 6 + 4\sqrt 6 \\ = 5 + 2\sqrt 6 = {\left( {\sqrt 2 + \sqrt 3 } \right)^2} \Rightarrow \sqrt {\Delta '} = \sqrt 2 + \sqrt 3 \\ \Rightarrow \cos x = \frac{{\sqrt 3 - \sqrt 2 + \sqrt 2 + \sqrt 3 }}{4} = \frac{{2\sqrt 3 }}{4} = \frac{{\sqrt 3 }}{2}\\ \cos x = \frac{{\sqrt 3 - \sqrt 2 - \sqrt 2 - \sqrt 3 }}{4} = \frac{{ - 2\sqrt 2 }}{4} = \frac{{ - \sqrt 2 }}{2}\\ \Rightarrow \left[ \begin{array}{l} \cos x = \frac{{\sqrt 3 }}{2}\\ \cos x = \frac{{ - \sqrt 2 }}{2} \end{array} \right. \Rightarrow \left[ \begin{array}{l} x = \pm \frac{\pi }{3} + k2\pi \\ x = \pm \frac{{3\pi }}{4} + k2\pi \end{array} \right.\left( {k \in Z} \right) \end{array}$
d) Điều kiện xác định $\cos x\ne 0\Rightarrow x\ne \dfrac{\pi} 2+k\pi$
$\begin{array}{l} 3{\tan ^2}x - 2\sqrt 3 \tan x + 3 = 0\\ \Leftrightarrow \Delta ' = b{'^2} - ac = 3 - 3.3 < 0 \end{array}$
Vậy phương trình vô nghiệm
