\[\begin{array}{l}
{\rm{a) - 2cos2x - 9sinx + 7 = 0}}\\
\Leftrightarrow - 2(1 - 2{\sin ^2}x) - 9\sin x + 7 = 0\\
\Leftrightarrow 4{\sin ^2}x - 9\sin x + 5 = 0\\
\Leftrightarrow 4{\sin ^2}x - 4\sin x - 5\sin x + 5 = 0\\
\Leftrightarrow 4\sin x(\sin x - 1) - 5(\sin x - 1) = 0\\
\Leftrightarrow (\sin x - 1)(4\sin x - 5) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 1\\
\sin x = \frac{5}{4}(ktm;do:\sin x \le 1)
\end{array} \right.\\
\Leftrightarrow x = \frac{\pi }{2} + k2\pi (k \in Z)\\
b)3si{n^2}2x - 2si{n^2}x = 0\\
\Leftrightarrow 3{(2\sin x\cos x)^2} - 2{\sin ^2}x = 0\\
\Leftrightarrow 12{\sin ^2}x{\cos ^2}x - 2{\sin ^2}x = 0\\
\Leftrightarrow 2{\sin ^2}x(6{\cos ^2}x - 1) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
6{\cos ^2}x - 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
\cos x = \frac{1}{{\sqrt 6 }}\\
\cos x = \frac{{ - 1}}{{\sqrt 6 }}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \pm \arccos \frac{1}{{\sqrt 6 }} + k2\pi \\
x = \pm \arccos \frac{{ - 1}}{{\sqrt 6 }} + k2\pi
\end{array} \right.(k \in Z)
\end{array}\]