Giải thích các bước giải:
Ta có:
\[\begin{array}{l}
1,\\
\cos 7x.\cos x = \cos 5x.\cos 3x\\
\Leftrightarrow \frac{1}{2}\left[ {\cos \left( {7x + x} \right) + \cos \left( {7x - x} \right)} \right] = \frac{1}{2}\left[ {\cos \left( {5x + 3x} \right) + \cos \left( {5x - 3x} \right)} \right]\\
\Leftrightarrow \cos 8x + \cos 6x = \cos 8x + \cos 2x\\
\Leftrightarrow \cos 6x = \cos 2x\\
\Leftrightarrow \left[ \begin{array}{l}
6x = 2x + k2\pi \\
6x = - 2x + k2\pi
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = \frac{{k\pi }}{2}\\
x = \frac{{k\pi }}{4}
\end{array} \right. \Rightarrow x = \frac{{k\pi }}{4}\\
2,\\
{\sin ^4}x + {\cos ^4}x = 1\\
\Leftrightarrow {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x.co{s^2}x = 1\\
\Leftrightarrow {1^2} - 2{\sin ^2}x.{\cos ^2}x = 1\\
\Leftrightarrow \sin x.{\mathop{\rm cosx}\nolimits} = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\cos x = 0
\end{array} \right. \Rightarrow x = \frac{{k\pi }}{2}
\end{array}\]
