Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\sin 3x = - \sin x\\
\Leftrightarrow \sin 3x = \sin \left( { - x} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
3x = - x + k2\pi \\
3x = \pi - \left( { - x} \right) + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{2}\\
x = \dfrac{\pi }{2} + k\pi
\end{array} \right. \Leftrightarrow x = \dfrac{{k\pi }}{2}\\
b,\\
\cos 2x = - \cos x\\
\Leftrightarrow \cos 2x = \cos \left( {\pi - x} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \pi - x + k2\pi \\
2x = x - \pi + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{3} + \dfrac{{k2\pi }}{3}\\
x = - \pi + k2\pi
\end{array} \right.\\
c,\\
\cos 3x = \sin 2x\\
\Leftrightarrow \cos 3x = \cos \left( {\dfrac{\pi }{2} - 2x} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
3x = \dfrac{\pi }{2} - 2x + k2\pi \\
3x = 2x - \dfrac{\pi }{2} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{10}} + \dfrac{{k2\pi }}{5}\\
x = - \dfrac{\pi }{2} + k2\pi
\end{array} \right.\\
d,\\
\cos 7x = - \sin 3x\\
\Leftrightarrow \cos 7x = \sin \left( { - 3x} \right)\\
\Leftrightarrow \cos 7x = \cos \left( {\dfrac{\pi }{2} + 3x} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
7x = \dfrac{\pi }{2} + 3x + k2\pi \\
7x = - \dfrac{\pi }{2} - 3x + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\\
x = - \dfrac{\pi }{{20}} + \dfrac{{k\pi }}{5}
\end{array} \right.\\
e,\\
\sin \left( {2x + \dfrac{\pi }{6}} \right) = - \cos \left( {x + \dfrac{\pi }{3}} \right)\\
\Leftrightarrow - \sin \left( { - 2x - \dfrac{\pi }{6}} \right) = - \cos \left( {x + \dfrac{\pi }{3}} \right)\\
\Leftrightarrow \sin \left( { - 2x - \dfrac{\pi }{6}} \right) = \cos \left( {x + \dfrac{\pi }{3}} \right)\\
\Leftrightarrow \cos \left( {\dfrac{\pi }{2} - \left( { - 2x - \dfrac{\pi }{6}} \right)} \right) = \cos \left( {x + \dfrac{\pi }{3}} \right)\\
\Leftrightarrow \cos \left( {\dfrac{{2\pi }}{3} + 2x} \right) = \cos \left( {x + \dfrac{\pi }{3}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{{2\pi }}{3} + 2x = x + \dfrac{\pi }{3} + k2\pi \\
\dfrac{{2\pi }}{3} + 2x = - x - \dfrac{\pi }{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - \dfrac{\pi }{3} + k2\pi \\
x = - \dfrac{\pi }{3} + \dfrac{{k2\pi }}{3}
\end{array} \right.\\
f,\\
\cos \left( {3x - \dfrac{\pi }{4}} \right) = \sin \left( {x - \dfrac{\pi }{3}} \right)\\
\Leftrightarrow \cos \left( {3x - \dfrac{\pi }{4}} \right) = \cos \left( {\dfrac{\pi }{2} - \left( {x - \dfrac{\pi }{3}} \right)} \right)\\
\Leftrightarrow \cos \left( {3x - \dfrac{\pi }{4}} \right) = \cos \left( {\dfrac{{5\pi }}{6} - x} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
3x - \dfrac{\pi }{4} = \dfrac{{5\pi }}{6} - x + k2\pi \\
3x - \dfrac{\pi }{4} = x - \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{13\pi }}{{48}} + \dfrac{{k\pi }}{2}\\
x = - \dfrac{{7\pi }}{{24}} + k\pi
\end{array} \right.
\end{array}\)
