Giải phương trình \(8\cos x = \frac{{\sqrt 3 }}{{\...

A

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

ĐK: \(\left\{ \begin{array}{l}\sin x \ne 0\\cosx \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ne k\pi \\x \ne \frac{\pi }{2} + k\pi \end{array} \right. \Leftrightarrow x \ne \frac{{k\pi }}{2}\,\,\left( {k \in \mathbb{Z}} \right)\).

\(8\cos x = \frac{{\sqrt 3 }}{{\sin x}} + \frac{1}{{\cos x}} \Leftrightarrow 8{\cos ^2}x.\sin x = \left( {\sqrt 3 .\cos x + \sin x} \right)\)

Chia cả 2 vế cho \({\cos ^3}x\) ta có:

\(\begin{array}{l}{\rm{     }}8.\frac{{\sin x}}{{\cos x}} = \frac{{\sqrt 3 }}{{{{\cos }^2}x}} + \frac{{\sin x}}{{{{\cos }^3}x}}\\ \Leftrightarrow 8\tan x = \sqrt 3 \left( {{{\tan }^2}x + 1} \right) + \tan x\left( {ta{n^2}x + 1} \right)\\ \Leftrightarrow 8\tan x = \sqrt 3 {\tan ^2}x + \sqrt 3  + {\tan ^3}x + \tan x\\ \Leftrightarrow {\tan ^3}x + \sqrt 3 {\tan ^2}x - 7\tan x + \sqrt 3  = 0\\ \Leftrightarrow \left[ \begin{array}{l}\tan x = 2 - \sqrt 3 \\\tan x =  - 2 - \sqrt 3 \\\tan x = \sqrt 3 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x =  - \frac{{11\pi }}{{12}} + k\pi \\x = \frac{{ - 5\pi }}{{12}} + k\pi \\x = \frac{{ - 2\pi }}{3} + k\pi \end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\,\,\left( {tm} \right)\end{array}\)

Chọn A.