Đáp án:
\[\left[ \begin{array}{l}
x = - \dfrac{\pi }{4} + k\pi \\
x = k2\pi
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
ĐKXĐ: \(\cos x \ne 0 \Leftrightarrow x \ne \dfrac{\pi }{2} + k\pi \)
Ta có:
\(\begin{array}{l}
1 + \tan x = \sin x + \cos x\\
\Leftrightarrow 1 + \dfrac{{\sin x}}{{\cos x}} = \sin x + \cos x\\
\Leftrightarrow \dfrac{{\cos x + \sin x}}{{\cos x}} = \sin x + \cos x\\
\Leftrightarrow \left( {\sin x + \cos x} \right)\left( {\dfrac{1}{{\cos x}} - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x + \cos x = 0\\
\dfrac{1}{{\cos x}} = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) = 0\\
\cos x = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin \left( {x + \dfrac{\pi }{4}} \right) = 0\\
\cos x = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{4} = k\pi \\
x = k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - \dfrac{\pi }{4} + k\pi \\
x = k2\pi
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
