Đáp án:

c. \(\left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x =  - \dfrac{\pi }{4} + k\pi 
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
a.{\sin ^2}x = {\cos ^2}x\\
 \to {\cos ^2}x - {\sin ^2}x = 0\\
 \to \cos 2x = 0\\
 \to 2x = \dfrac{\pi }{2} + k\pi \\
 \to x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\left( {k \in Z} \right)\\
b.2\sin x.\cos x.\cos 2x = 1\\
 \to \sin 2x.\cos 2x = 1\\
 \to \dfrac{{\sin 4x}}{2} = 1\\
 \to \sin 4x = 2\left( {vô lý} \right)\\
 \to x \in \emptyset \\
c.\sin x.\cos x.\sin 2x = \dfrac{1}{2}\\
 \to \dfrac{{\sin 2x}}{2}.\sin 2x = \dfrac{1}{2}\\
 \to {\sin ^2}2x = 1\\
 \to \left[ \begin{array}{l}
\sin 2x = 1\\
\sin 2x =  - 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
2x = \dfrac{\pi }{2} + k2\pi \\
2x =  - \dfrac{\pi }{2} + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
 \to \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x =  - \dfrac{\pi }{4} + k\pi 
\end{array} \right.\left( {k \in Z} \right)\\
d.{\sin ^4}x + {\cos ^4}x =  - \dfrac{1}{2}.{\cos ^2}2x\\
 \to \left( {{{\sin }^4}x + {{\cos }^4}x + 2{{\sin }^2}x.{{\cos }^2}x} \right) - 2{\sin ^2}x{\cos ^2}x =  - \dfrac{1}{2}.{\cos ^2}2x\\
 \to {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x{\cos ^2}x =  - \dfrac{1}{2}.{\cos ^2}2x\\
 \to 1 - 2{\left( {\sin x.\cos x} \right)^2} =  - \dfrac{1}{2}.{\cos ^2}2x\\
 \to 1 - 2{\left( {\dfrac{{\sin 2x}}{2}} \right)^2} =  - \dfrac{1}{2}.{\cos ^2}2x\\
 \to 1 - \dfrac{{{{\sin }^2}2x}}{2} =  - \dfrac{{{{\cos }^2}2x}}{2}\\
 \to \dfrac{{{{\cos }^2}2x}}{2} - \dfrac{{{{\sin }^2}2x}}{2} + 1 = 0\\
 \to {\cos ^2}2x - {\sin ^2}2x =  - 2\\
 \to \cos 2.2x =  - 2\\
 \to \cos 4x =  - 2\left( {vô lý} \right)\\
 \to x \in \emptyset 
\end{array}\)