Đáp án:
\(\begin{array}{l}
a)\,\,x = \pm \frac{\pi }{6} + k\pi \,\,\left( {k \in Z} \right)\\
b)\,\,x = \pi + k2\pi \,\,\left( {k \in Z} \right)
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\,\,{\sin ^2}2x - 2{\cos ^2}x + \frac{3}{4} = 0\\
\Leftrightarrow 1 - {\cos ^2}2x - \left( {1 + \cos 2x} \right) + \frac{3}{4} = 0\\
\Leftrightarrow - {\cos ^2}2x - \cos 2x + \frac{3}{4} = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 2x = \frac{1}{2}\\
\cos 2x = - \frac{3}{2}\,\,\left( {loai} \right)
\end{array} \right.\\
\Leftrightarrow 2x = \pm \frac{\pi }{3} + k2\pi \\
\Leftrightarrow x = \pm \frac{\pi }{6} + k\pi \,\,\left( {k \in Z} \right)\\
b)\,\,\cos 2x + {\sin ^2}x + 2\cos x + 1 = 0\\
\Leftrightarrow {\cos ^2}x - {\sin ^2}x + 2\cos x + 1 = 0\\
\Leftrightarrow {\cos ^2}x + 2\cos x + 1 = 0\\
\Leftrightarrow {\left( {\cos x + 1} \right)^2} = 0\\
\Leftrightarrow \cos x = - 1\\
\Leftrightarrow x = \pi + k2\pi \,\,\left( {k \in Z} \right)
\end{array}\)
