Đáp án:
x. \(\left[ \begin{array}{l}
x = \dfrac{\pi }{{16}} + \dfrac{b}{8} + \dfrac{{k\pi }}{4}\\
x = - \dfrac{\pi }{4} - \dfrac{b}{2} + k\pi
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
w.\cos \left( {2x - a} \right) = \sin 3x\\
\to \cos \left( {2x - a} \right) = \cos \left( {\dfrac{\pi }{2} - 3x} \right)\\
\to \left[ \begin{array}{l}
2x - a = \dfrac{\pi }{2} - 3x + k2\pi \\
2x - a = - \dfrac{\pi }{2} + 3x + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
5x = \dfrac{\pi }{2} + a + k2\pi \\
x = \dfrac{\pi }{2} - a + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{{10}} + \dfrac{a}{5} + \dfrac{{k2\pi }}{5}\\
x = \dfrac{\pi }{2} - a + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
x.\sin \left( {3x - b} \right) = \cos 5x\\
\to \sin \left( {3x - b} \right) = \sin \left( {\dfrac{\pi }{2} - 5x} \right)\\
\to \left[ \begin{array}{l}
3x - b = \dfrac{\pi }{2} - 5x + k2\pi \\
3x - b = \pi - \dfrac{\pi }{2} + 5x + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
8x = \dfrac{\pi }{2} + b + k2\pi \\
2x = - \dfrac{\pi }{2} - b + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{{16}} + \dfrac{b}{8} + \dfrac{{k\pi }}{4}\\
x = - \dfrac{\pi }{4} - \dfrac{b}{2} + k\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)
