Đáp án:
$\begin{array}{l}
a)\sin x - \frac{1}{2} = 0\\
\Rightarrow \sin x = \frac{1}{2}\\
\Rightarrow \left[ \begin{array}{l}
x = \frac{\pi }{6} + k2\pi \\
x = \pi - \frac{\pi }{6} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
\Rightarrow \left[ \begin{array}{l}
x = \frac{\pi }{6} + k2\pi \\
x = \frac{{5\pi }}{6} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
b)2{\cos ^2}x + 5\cos x - 7 = 0\\
\Rightarrow 2{\cos ^2}x - 2{\mathop{\rm cosx}\nolimits} + 7cosx - 7 = 0\\
\Rightarrow \left( {\cos x - 1} \right)\left( {2\cos x + 7} \right) = 0\\
\Rightarrow \cos x = 1\left( {do: - 1 \le \cos x \le 1} \right)\\
\Rightarrow x = k2\pi \left( {k \in Z} \right)\\
c)2\sqrt 3 \sin x + 3\sqrt 3 \tan x = 2\cos x + 3\\
\left( {dk:x \ne \frac{\pi }{2} + k\pi } \right)\\
\Rightarrow 2\sqrt 3 \sin x + \frac{{3\sqrt 3 \sin x}}{{\cos x}} = 2\cos x + 3\\
\Rightarrow \frac{{\sqrt 3 \sin x\left( {2\cos x + 3} \right)}}{{\cos x}} = 2\cos x + 3\\
\Rightarrow \frac{{\sqrt 3 \sin x}}{{\cos x}} = 1\left( {do:2\cos x + 3 > 0\forall x} \right)\\
\Rightarrow \tan x = \frac{1}{{\sqrt 3 }}\\
\Rightarrow x = \frac{\pi }{6} + k\pi
\end{array}$
