a,
ĐK: $\cos x\ne 0$
$\to x\ne \dfrac{\pi}{2}+k\pi$
$\dfrac{2}{\cos^2x}+3\tan x-9=0$
$\to 2+2\tan^2x+3\tan x-9=0$
$\to 2\tan^2x+3\tan x-7=0$
$\to \tan x=\dfrac{-3\pm\sqrt{65}}{4}$
$\to x=\arctan\dfrac{-3\pm\sqrt{65}}{4}+k\pi$ (TM)
b,
$3\cos2x-4\sin2x=1$
$\to 5\cos(2x+\alpha)=1$
Với $\begin{cases} \cos\alpha=\dfrac{3}{5}\\ \sin\alpha=\dfrac{4}{5}\end{cases}$
$\to \cos(2x+\alpha)=\dfrac{1}{5}$
$\to \left[ \begin{array}{l}x=\dfrac{-\alpha}{2}+\arccos\dfrac{1}{5}+k\pi \\x=\dfrac{-\alpha}{2}+\dfrac{\pi}{2}-\dfrac{1}{2}\arccos\dfrac{1}{5}+k\pi \end{array} \right.$
