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Đáp án:

 b) $\left[ \begin{array}{l} x = k2\pi \\ x = \arccos \dfrac{1}{6} + k2\pi \\ x =  - \arccos \dfrac{1}{6} + k2\pi  \end{array} \right.$

Giải thích các bước giải:

$\begin{array}{l} a)4{\cos ^2}x - 2\cos x - 2 = 0\\  \to 2\left( {2\cos x + 1} \right)\left( {\cos x - 1} \right) = 0\\  \to \left[ \begin{array}{l} \cos x =  - \dfrac{1}{2}\\ \cos x = 1 \end{array} \right.\\  \to \left[ \begin{array}{l} x = \dfrac{{2\pi }}{3} + k2\pi \\ x =  - \dfrac{{2\pi }}{3} + k2\pi \\ x = k2\pi  \end{array} \right.\left( {k \in Z} \right)\\ b)6{\sin ^2}x + 7\cos x - 7 = 0\\  \to 6\left( {1 - {{\cos }^2}x} \right) + 7\cos x - 7 = 0\\  \to  - 6{\cos ^2}x + 7\cos x - 1 = 0\\  \to \left( {1 - \cos x} \right)\left( {6\cos x - 1} \right) = 0\\  \to \left[ \begin{array}{l} \cos x = 1\\ \cos x = \dfrac{1}{6} \end{array} \right.\\  \to \left[ \begin{array}{l} x = k2\pi \\ x = \arccos \dfrac{1}{6} + k2\pi \\ x =  - \arccos \dfrac{1}{6} + k2\pi  \end{array} \right.\left( {k \in Z} \right)\\ c)2{\sin ^2}x - 4\cos x\sin x - 3{\cos ^2}x =  - 2\\  \to \dfrac{{2{{\sin }^2}x}}{{{{\cos }^2}x}} - \dfrac{{4\sin x}}{{\cos x}} - 3 =  - \dfrac{2}{{{{\cos }^2}x}}\\  \to 2{\tan ^2}x - 4\tan x - 3 =  - 2\left( {1 + {{\tan }^2}x} \right)\\  \to 4{\tan ^2}x - 4\tan x - 1 = 0\\  \to \left[ \begin{array}{l} \tan x = \dfrac{{1 + \sqrt 2 }}{2}\\ \tan x = \dfrac{{1 - \sqrt 2 }}{2} \end{array} \right.\\  \to \left[ \begin{array}{l} x = \arctan \left( {\dfrac{{1 + \sqrt 2 }}{2}} \right) + k\pi \\ x = \arctan \left( {\dfrac{{1 - \sqrt 2 }}{2}} \right) + k\pi  \end{array} \right.\left( {k \in Z} \right) \end{array}$