Đáp án:
\[x = \pm \dfrac{\pi }{6} + k\pi \,\,\,\,\,\,\,\,\,\,\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
4\left( {{{\sin }^4}x + {{\cos }^4}x} \right) = 5.\cos 2x\\
\Leftrightarrow 4.\left[ {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} - 2.{{\sin }^2}x.{{\cos }^2}x} \right] = 5\cos 2x\\
\Leftrightarrow 4.\left[ {{1^2} - \dfrac{1}{2}.{{\left( {2\sin x.\cos x} \right)}^2}} \right] = 5\cos 2x\\
\Leftrightarrow 4.\left( {1 - \dfrac{1}{2}{{\sin }^2}2x} \right) = 5\cos 2x\\
\Leftrightarrow 4 - 2{\sin ^2}2x = 5\cos 2x\\
\Leftrightarrow 4 - 2.\left( {1 - {{\cos }^2}2x} \right) = 5\cos 2x\\
\Leftrightarrow 2 + 2{\cos ^2}2x = 5\cos 2x\\
\Leftrightarrow 2{\cos ^2}2x - 5\cos 2x + 2 = 0\\
\Leftrightarrow \left( {\cos 2x - 2} \right)\left( {2\cos 2x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 2x = 2\\
\cos 2x = \dfrac{1}{2}
\end{array} \right.\\
- 1 \le \cos 2x \le 1\\
\Rightarrow \cos 2x = \dfrac{1}{2}\\
\Leftrightarrow 2x = \pm \dfrac{\pi }{3} + k2\pi \\
\Leftrightarrow x = \pm \dfrac{\pi }{6} + k\pi \,\,\,\,\,\,\,\,\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
