tanx + cosx – 2=0 giup minh ak

Đáp án:

\(x = \dfrac{\pi }{4} + k\pi \)

Giải thích các bước giải:

\(\begin{array}{l}
DK:\left\{ \begin{array}{l}
\sin x \ne 0\\
\cos x \ne 0
\end{array} \right. \to \sin 2x \ne 0\\
 \to 2x \ne k\pi  \to x \ne \dfrac{{k\pi }}{2}\left( {k \in Z} \right)\\
\tan x + \cot x – 2 = 0\\
 \to \tan x + \dfrac{1}{{\tan x}} – 2 = 0\\
 \to {\tan ^2}x – 2\tan x + 1 = 0\\
 \to {\left( {\tan x – 1} \right)^2} = 0\\
 \to \tan x – 1 = 0\\
 \to \tan x = 1\\
 \to x = \dfrac{\pi }{4} + k\pi \left( {k \in Z} \right)
\end{array}\)