Đáp án:

\(\begin{array}{l}
a,\\
x = \dfrac{\pi }{4} + k\pi \,\,\,\,\left( {k \in Z} \right)\\
b,\\
x = \dfrac{\pi }{4} + k\pi \,\,\,\,\left( {k \in Z} \right)\\
c,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x =  - \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
d,\\
x = \dfrac{{k\pi }}{2}\,\,\,\,\left( {k \in Z} \right)
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
\sin x = \cos x\\
 \Leftrightarrow \sin x = \sin \left( {\dfrac{\pi }{2} - x} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} - x + k2\pi \\
x = \pi  - \left( {\dfrac{\pi }{2} - x} \right) + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{2} + k2\pi \\
x = \dfrac{\pi }{2} + x + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
0x = \dfrac{\pi }{2} + k2\pi \,\,\,\left( L \right)
\end{array} \right.\\
 \Leftrightarrow x = \dfrac{\pi }{4} + k\pi \,\,\,\,\left( {k \in Z} \right)\\
b,\\
\sin x - \cos x = 0\\
 \Leftrightarrow \sin x = \cos x\\
 \Leftrightarrow \sin x = \sin \left( {\dfrac{\pi }{2} - x} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} - x + k2\pi \\
x = \pi  - \left( {\dfrac{\pi }{2} - x} \right) + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{2} + k2\pi \\
x = \dfrac{\pi }{2} + x + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
0x = \dfrac{\pi }{2} + k2\pi \,\,\,\left( L \right)
\end{array} \right.\\
 \Leftrightarrow x = \dfrac{\pi }{4} + k\pi \,\,\,\,\left( {k \in Z} \right)\\
c,\\
\sin x + \cos 3x = 0\\
 \Leftrightarrow \cos 3x =  - \sin x\\
 \Leftrightarrow \cos 3x = \sin \left( { - x} \right)\\
 \Leftrightarrow \cos 3x = \cos \left( {\dfrac{\pi }{2} - \left( { - x} \right)} \right)\\
 \Leftrightarrow \cos 3x = \cos \left( {\dfrac{\pi }{2} + x} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
3x = \dfrac{\pi }{2} + x + k2\pi \\
3x =  - \dfrac{\pi }{2} - x + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{2} + k2\pi \\
4x =  - \dfrac{\pi }{2} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x =  - \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
d,\\
\sin x.\cos x = 0\\
 \Leftrightarrow \dfrac{1}{2}.\left( {2\sin x.\cos x} \right) = 0\\
 \Leftrightarrow \dfrac{1}{2}\sin 2x = 0\\
 \Leftrightarrow \sin 2x = 0\\
 \Leftrightarrow 2x = k\pi \\
 \Leftrightarrow x = \dfrac{{k\pi }}{2}\,\,\,\,\left( {k \in Z} \right)
\end{array}\)