Giải phương trình \(\left( {1 + \tan x} \right)\le...

A

- Hướng dẫn giải

Lời giải chi tiết:

ĐK: \(\cos x \ne 0 \Leftrightarrow x \ne \frac{\pi }{2} + k\pi \,\,\left( {k \in \mathbb{Z}} \right)\).

\(\) \(\begin{array}{l}\,\,\,\,\,\,\left( {1 + \tan x} \right)\left( {1 + \sin 2x} \right) = 1 + \tan x.\\ \Leftrightarrow \left( {1 + \tan x} \right)\left( {1 + \sin 2x} \right) - \left( {1 + \tan x} \right) = 0\\ \Leftrightarrow \left( {1 + \tan x} \right)\left( {1 + \sin 2x - 1} \right) = 0\\ \Leftrightarrow \left( {1 + \tan x} \right).sin2x = 0\\ \Leftrightarrow \left[ \begin{array}{l}1 + \tan x = 0\\\sin 2x = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\tan x =  - 1\\\sin 2x = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{{ - \pi }}{4} + k\pi \\2x = k\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{{ - \pi }}{4} + k\pi \,\,\left( {tm} \right)\\x = \frac{{k\pi }}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {tm} \right)\end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

Vậy \(S = \left\{ { - \frac{\pi }{4} + k\pi ,\,\,\frac{{k\pi }}{2},\,\,k \in \mathbb{Z}} \right\}\).

Chọn A.