Đáp án:
$\left[ \begin{array}{l}
x = k\pi \\
x = \frac{\pi }{4} + k2\pi \\
x = \frac{{3\pi }}{4} + k2\pi \\
x = - \frac{\pi }{4} + k2\pi \\
x = \frac{{5\pi }}{4} + k2\pi
\end{array} \right.\,\,\,\left( {k \in Z} \right).$
Giải thích các bước giải:
$\begin{array}{l}
\,\,\,\,\,\,\,\,\,2{\cos ^4}x + 3{\sin ^2}x - 2 = 0\\
\Leftrightarrow 2{\left( {1 - {{\sin }^2}x} \right)^2} + 3{\sin ^2}x - 2 = 0\\
\Leftrightarrow 2\left( {1 - 2{{\sin }^2}x + {{\sin }^4}x} \right) + 3{\sin ^2}x - 2 = 0\\
\Leftrightarrow 2{\sin ^4}x - 4{\sin ^2}x + 2 + 3{\sin ^2}x - 2 = 0\\
\Leftrightarrow 2{\sin ^4}x - {\sin ^2}x = 0\\
\Leftrightarrow {\sin ^2}x\left( {2{{\sin }^2}x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{\sin ^2}x = 0\\
{\sin ^2}x = \frac{1}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\sin x = \frac{1}{{\sqrt 2 }}\\
\sin x = - \frac{1}{{\sqrt 2 }}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \frac{\pi }{4} + k2\pi \\
x = \pi - \frac{\pi }{4} + k2\pi \\
x = - \frac{\pi }{4} + k2\pi \\
x = \pi + \frac{\pi }{4} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \frac{\pi }{4} + k2\pi \\
x = \frac{{3\pi }}{4} + k2\pi \\
x = - \frac{\pi }{4} + k2\pi \\
x = \frac{{5\pi }}{4} + k2\pi
\end{array} \right.\,\,\,\left( {k \in Z} \right).
\end{array}$
