hộ mình với ạ mình cảm ơn Question hộ mình với ạ mình cảm ơn in progress 0 Môn Toán Tryphena 4 years 2020-10-24T09:49:23+00:00 2020-10-24T09:49:23+00:00 1 Answers 76 views 0
Answers ( )
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:
\(\begin{array}{l}
2.\left( {\sqrt 3 \sin x – \cos x} \right) = – 3.\left( {\sqrt 3 \tan x – 1} \right)\\
\Leftrightarrow 2.\left( {\sqrt 3 \sin x – \cos x} \right) = – 3.\left( {\dfrac{{\sqrt 3 \sin x}}{{\cos x}} – 1} \right)\\
\Leftrightarrow 2.\left( {\sqrt 3 \sin x – \cos x} \right) = – 3.\dfrac{{\sqrt 3 \sin x – \cos x}}{{\cos x}}\\
\Leftrightarrow \left( {\sqrt 3 \sin x – \cos x} \right).\left( {2 + \dfrac{3}{{\cos x}}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt 3 \sin x – \cos x = 0\\
2 + \dfrac{3}{{\cos x}} = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt 3 \sin x = \cos x\\
\cos x = – \dfrac{3}{2}\,\,\,\,\,\left( {L,\,\,\, – 1 \le \cos x \le 1} \right)
\end{array} \right.\\
\Leftrightarrow \dfrac{{\sin x}}{{\cos x}} = \dfrac{1}{{\sqrt 3 }}\,\,\,\,\,\,\,\,\,\,\,\left( {\cos x \ne 0} \right)\\
\Leftrightarrow \tan x = \dfrac{1}{{\sqrt 3 }}\\
\Leftrightarrow x = \dfrac{\pi }{6} + k\pi \,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)