Phương trình \(\cos x + \sqrt 3 \sin x = 3 - \dfrac{3}{{\cos x + \sqrt 3 \sin x + 1}} \) có nghiệm là:

Bước 1:

ĐK:

\(\begin{array}{l}\,\,\,\,\,\,\cos x + \sqrt 3 \sin x + 1 \ne 0\\ \Leftrightarrow 2\left( {\dfrac{{\sqrt 3 }}{2}\sin x + \dfrac{1}{2}\cos x} \right) \ne  - 1\\ \Leftrightarrow 2\left( {\sin x\cos \dfrac{\pi }{6} + \cos x\sin \dfrac{\pi }{6}} \right) \ne  - 1\\\, \Leftrightarrow 2\sin \left( {x + \dfrac{\pi }{6}} \right) \ne  - 1\\ \Leftrightarrow \sin \left( {x + \dfrac{\pi }{6}} \right) \ne  - \dfrac{1}{2} = \sin \left( {\dfrac{{ - \pi }}{6}} \right)\\ \Leftrightarrow \left\{ \begin{array}{l}x + \dfrac{\pi }{6} \ne \dfrac{{ - \pi }}{6} + k2\pi \\x + \dfrac{\pi }{6} \ne \pi  - \dfrac{{ - \pi }}{6} + k2\pi \end{array} \right. \\\Leftrightarrow \left\{ \begin{array}{l}x \ne \dfrac{{ - \pi }}{3} + k2\pi \\x \ne \pi  + k2\pi \end{array} \right.\end{array}\)

Bước 2:

Đặt \(t = \cos x + \sqrt 3 \sin x\)\( = 2\left( {\dfrac{{\sqrt 3 }}{2}\sin x + \dfrac{1}{2}\cos x} \right)\)\( = 2\left( {\sin x\cos \dfrac{\pi }{6} + \cos x\sin \dfrac{\pi }{6}} \right)\)\( = 2\sin \left( {x + \dfrac{\pi }{6}} \right) \in \left[ { - 2;2} \right]\backslash \left\{ { - 1} \right\}\)

Bước 3:

Khi đó phương trình đã cho

$\begin{array}{l} \Leftrightarrow t = 3 - \dfrac{3}{{t + 1}} \Leftrightarrow t = \dfrac{{3t + 3}}{{t + 1}} - \dfrac{3}{{t + 1}}\\ \Leftrightarrow {t^2} + t = 3t + 3 - 3 \Leftrightarrow {t^2} - 2t = 0 \\\Leftrightarrow \left[ \begin{array}{l}t = 0\,\,\left( {tm} \right)\\t = 2\,\,\,\left( {tm} \right)\end{array} \right.\end{array}$

Bước 4:

$\Leftrightarrow \left[ \begin{array}{l}2\sin \left( {x + \dfrac{\pi }{6}} \right) = 0\\2\sin \left( {x + \dfrac{\pi }{6}} \right) = 2\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\sin \left( {x + \dfrac{\pi }{6}} \right) = 0\\\sin \left( {x + \dfrac{\pi }{6}} \right) = 1\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x + \dfrac{\pi }{6} = k\pi \\x + \dfrac{\pi }{6} = \dfrac{\pi }{2} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x =  - \dfrac{\pi }{6} + k\pi \\x = \dfrac{\pi }{3} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{{5\pi }}{6} + k\pi \\x = \dfrac{\pi }{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)$