## hộ mình với ạ mình cảm ơn

Question

hộ mình với ạ mình cảm ơn

in progress 0
4 years 2020-10-24T09:49:23+00:00 1 Answers 76 views 0

1. Đáp án:

$x = \dfrac{\pi }{6} + k\pi \,\,\,\,\,\left( {k \in Z} \right)$

Giải thích các bước giải:

ĐKXĐ:  $$\cos x \ne 0 \Leftrightarrow x \ne \dfrac{\pi }{2} + k\pi$$

Ta có:

$$\begin{array}{l} 2.\left( {\sqrt 3 \sin x – \cos x} \right) = – 3.\left( {\sqrt 3 \tan x – 1} \right)\\ \Leftrightarrow 2.\left( {\sqrt 3 \sin x – \cos x} \right) = – 3.\left( {\dfrac{{\sqrt 3 \sin x}}{{\cos x}} – 1} \right)\\ \Leftrightarrow 2.\left( {\sqrt 3 \sin x – \cos x} \right) = – 3.\dfrac{{\sqrt 3 \sin x – \cos x}}{{\cos x}}\\ \Leftrightarrow \left( {\sqrt 3 \sin x – \cos x} \right).\left( {2 + \dfrac{3}{{\cos x}}} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sqrt 3 \sin x – \cos x = 0\\ 2 + \dfrac{3}{{\cos x}} = 0 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} \sqrt 3 \sin x = \cos x\\ \cos x = – \dfrac{3}{2}\,\,\,\,\,\left( {L,\,\,\, – 1 \le \cos x \le 1} \right) \end{array} \right.\\ \Leftrightarrow \dfrac{{\sin x}}{{\cos x}} = \dfrac{1}{{\sqrt 3 }}\,\,\,\,\,\,\,\,\,\,\,\left( {\cos x \ne 0} \right)\\ \Leftrightarrow \tan x = \dfrac{1}{{\sqrt 3 }}\\ \Leftrightarrow x = \dfrac{\pi }{6} + k\pi \,\,\,\,\,\left( {k \in Z} \right) \end{array}$$