Đáp án:
b. \(x \ne \dfrac{{k\pi }}{2}\)
Giải thích các bước giải:
\(\begin{array}{*{20}{l}}
{a.y = \dfrac{{1 + \cos x}}{{\sqrt 3 + 2\sin 2x}}}\\
\begin{array}{l}
DK:\sqrt 3 + 2\sin 2x \ne 0\\
\to \sin 2x \ne - \dfrac{{\sqrt 3 }}{2}
\end{array}\\
\begin{array}{l}
\to \left[ \begin{array}{l}
2x \ne - \dfrac{\pi }{3} + k2\pi \\
2x \ne \dfrac{{4\pi }}{3} + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x \ne - \dfrac{\pi }{6} + k\pi \\
x \ne \dfrac{{2\pi }}{3} + k\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}\\
{b.y = \dfrac{1}{{\cos x}} + \dfrac{1}{{\sin x}}}\\
{DK:\left\{ {\begin{array}{*{20}{l}}
{\sin x \ne 0}\\
{\cos x \ne 0}
\end{array}} \right. \to \dfrac{{\sin 2x}}{2} \ne 0\left( {do:\dfrac{{\sin 2x}}{2} = \sin x.\cos x} \right)}\\
{ \to \sin 2x \ne 0}\\
{ \to 2x \ne k\pi }\\
{ \to x \ne \dfrac{{k\pi }}{2}\left( {k \in Z} \right)}
\end{array}\)
