Đáp án:
\(\left\{ \begin{array}{l}
x \ne \dfrac{{17\pi }}{{140}} + \dfrac{{k2\pi }}{7}\\
x \ne \dfrac{{7\pi }}{{20}} + k2\pi
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
y = \dfrac{{1 + \sin x}}{{\cos \left( {4x + \dfrac{{2\pi }}{5}} \right) + \cos \left( {3x - \dfrac{\pi }{4}} \right)}}\\
DK:\cos \left( {4x + \dfrac{{2\pi }}{5}} \right) + \cos \left( {3x - \dfrac{\pi }{4}} \right) \ne 0\\
\to 2\cos \left( {\dfrac{{7x + \dfrac{{3\pi }}{{20}}}}{2}} \right)\cos \left( {\dfrac{{x + \dfrac{{13\pi }}{{20}}}}{2}} \right) \ne 0\\
\to \left\{ \begin{array}{l}
\cos \left( {\dfrac{{7x + \dfrac{{3\pi }}{{20}}}}{2}} \right) \ne 0\\
\cos \left( {\dfrac{{x + \dfrac{{13\pi }}{{20}}}}{2}} \right) \ne 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{{7x + \dfrac{{3\pi }}{{20}}}}{2} \ne \dfrac{\pi }{2} + k\pi \\
\dfrac{{x + \dfrac{{13\pi }}{{20}}}}{2} \ne \dfrac{\pi }{2} + k\pi
\end{array} \right.\\
\to \left\{ \begin{array}{l}
7x + \dfrac{{3\pi }}{{20}} \ne \pi + k2\pi \\
x + \dfrac{{13\pi }}{{20}} \ne \pi + k2\pi
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ne \dfrac{{17\pi }}{{140}} + \dfrac{{k2\pi }}{7}\\
x \ne \dfrac{{7\pi }}{{20}} + k2\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)
