Đáp án:
\[\left[ \begin{array}{l}
x = \pm \dfrac{\pi }{3} + k2\pi \\
x = \pm \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
4{\sin ^2}2x + 8{\cos ^2}x - 5 = 0\\
\Leftrightarrow 4.{\left( {2\sin x.\cos x} \right)^2} + 8.{\cos ^2}x - 5 = 0\\
\Leftrightarrow 16.{\sin ^2}x.{\cos ^2}x + 8{\cos ^2}x - 5 = 0\\
\Leftrightarrow 16.\left( {1 - {{\cos }^2}x} \right).{\cos ^2}x + 8{\cos ^2}x - 5 = 0\\
\Leftrightarrow 16{\cos ^2}x - 16{\cos ^4}x + 8{\cos ^2}x - 5 = 0\\
\Leftrightarrow - 16{\cos ^4}x + 24{\cos ^2}x - 5 = 0\\
\Leftrightarrow 16{\cos ^4}x - 24{\cos ^2}x + 5 = 0\\
\Leftrightarrow \left( {4{{\cos }^2}x - 1} \right)\left( {4{{\cos }^2}x - 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{\cos ^2}x = \dfrac{1}{4}\\
{\cos ^2}x = \dfrac{5}{4}
\end{array} \right.\\
- 1 \le \cos x \le 1 \Rightarrow 0 \le {\cos ^2}x \le 1\\
\Rightarrow {\cos ^2}x = \dfrac{1}{4} \Leftrightarrow \left[ \begin{array}{l}
\cos x = \dfrac{1}{2}\\
\cos x = - \dfrac{1}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \pm \dfrac{\pi }{3} + k2\pi \\
x = \pm \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)
