Đáp án:

d) \(\left[ \begin{array}{l}
x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\
x = \dfrac{\pi }{{12}} + k\pi \\
x = \dfrac{{5\pi }}{{12}} + k\pi 
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
1.6a)\\
\cos x\left( {\cos x - \sqrt 3 \sin x} \right) = 0\\
 \to \left[ \begin{array}{l}
\cos x = 0\\
\cos x - \sqrt 3 \sin x = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
\dfrac{1}{2}\cos x - \dfrac{{\sqrt 3 }}{2}\sin x = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
\sin \dfrac{\pi }{6}.\cos x - \cos \dfrac{\pi }{6}.\sin x = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
\sin \left( {\dfrac{\pi }{6} - x} \right) = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
\dfrac{\pi }{6} - x = k\pi 
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x = \dfrac{\pi }{6} + k\pi 
\end{array} \right.\left( {k \in Z} \right)\\
c)4.\sin 2x.\cos 2x = \cos 8\left( {\dfrac{\pi }{{16}} - x} \right)\\
 \to 2\sin 4x = \cos 8\left( {\dfrac{\pi }{{16}} - x} \right)\\
 \to 2\sin 4x = \cos \left( {\dfrac{\pi }{2} - 8x} \right)\\
 \to 2\sin 4x = \sin \left( {8x} \right)\\
 \to 2\sin 4x = 2\sin 4x.\cos 4x\\
 \to \sin 4x = \sin 4x.\cos 4x\\
 \to \sin 4x\left( {1 - \cos 4x} \right) = 0\\
 \to \left[ \begin{array}{l}
\sin 4x = 0\\
\cos 4x = 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{4}\\
x = \dfrac{{k\pi }}{2}
\end{array} \right.\left( {k \in Z} \right)\\
 \to x = \dfrac{{k\pi }}{4}\left( {k \in Z} \right)\\
d){\left( { - \cos x} \right)^4} - {\sin ^4}x = \sin 4x\\
 \to {\cos ^4}x - {\sin ^4}x = \sin 4x\\
 \to \left( {{{\cos }^2}x - {{\sin }^2}x} \right)\left( {{{\cos }^2}x + {{\sin }^2}x} \right) = 2\sin 2x.\cos 2x\\
 \to \cos 2x.1 = 2\sin 2x.\cos 2x\\
 \to \cos 2x\left( {1 - 2\sin 2x} \right) = 0\\
 \to \left[ \begin{array}{l}
\cos 2x = 0\\
\sin 2x = \dfrac{1}{2}
\end{array} \right.\\
 \to \left[ \begin{array}{l}
2x = \dfrac{\pi }{2} + k\pi \\
2x = \dfrac{\pi }{6} + k2\pi \\
2x = \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\
x = \dfrac{\pi }{{12}} + k\pi \\
x = \dfrac{{5\pi }}{{12}} + k\pi 
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)