Đáp án:
$x = \dfrac{\pi }{4} + k\pi $
Giải thích các bước giải:
\[\begin{array}{l}
1.\\
1 - 2.co{s^2}x.{\sin ^2}x = \cos 4x\\
\Leftrightarrow \dfrac{1}{2}{\sin ^2}2x = 1 - \cos 4x\\
\Leftrightarrow \dfrac{1}{2}{\sin ^2}2x = 1 - \left( {1 - 2{{\sin }^2}2x} \right)\\
\Leftrightarrow {\sin ^2}2x = 4{\sin ^2}2x\\
\Leftrightarrow {\sin ^2}2x = 0 \Leftrightarrow \sin 2x = 0\\
\Leftrightarrow 2x = \dfrac{\pi }{2} + k2\pi \\
\Leftrightarrow x = \dfrac{\pi }{4} + k\pi \\
2.\\
\left( {{{\cos }^2}x - {{\sin }^2}x} \right).\left( {{{\cos }^2}x + {{\sin }^2}x} \right) = \cos 3x\\
\Leftrightarrow \cos 2x = \cos 3x\\
\Leftrightarrow \left[ \begin{array}{l}
2x = 3x + k2\pi \\
2x = - 3x + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = k2\pi \\
x = \dfrac{{k2\pi }}{5}
\end{array} \right.\\
3.\\
\left( {{{\cos }^2}x + {{\sin }^2}x} \right).\left( {{{\cos }^2}x - \cos x.sinx + si{n^2}x} \right) = \cos 4x\\
\Leftrightarrow 1 - \cos x.sinx = \cos 4x\\
\Leftrightarrow \cos x.\sin x = 1 - \cos 4x = 2{\sin ^2}2x\\
\Leftrightarrow \dfrac{1}{2}\sin 2x = 2{\sin ^2}2x\\
\Leftrightarrow \sin 2x.\left( {4\sin 2x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 2x = 0\\
\sin 2x = \dfrac{1}{4}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
\left[ \begin{array}{l}
2x = \arcsin \dfrac{1}{4} + k2\pi \\
2x = \pi - \arcsin \dfrac{1}{4} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{\arcsin \dfrac{1}{4}}}{2} + k\pi \\
x = \dfrac{{\pi - \arcsin \dfrac{1}{4}}}{2} + k2\pi
\end{array} \right.
\end{array} \right.
\end{array}\]
