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