Đáp án:
${x = \dfrac{\pi }{6} + k\pi \left( {k \in Z} \right);x = \dfrac{\pi }{3} + k2\pi \left( {k \in Z} \right);x = - \dfrac{\pi }{9} + k\dfrac{{2\pi }}{3}\left( {k \in Z} \right)}$
Giải thích các bước giải:
ĐKXĐ: $x \ne \dfrac{\pi }{4} + k\dfrac{\pi }{2}$
$\begin{array}{l}
\dfrac{{2 - \cos 2x - \sqrt 3 .\sin 2x}}{{2\cos 2x}} = \sqrt 3 \sin x - \cos x\\
\Leftrightarrow \dfrac{{2\left( {{{\cos }^2}x + {{\sin }^2}x} \right) - \left( {{{\cos }^2}x - {{\sin }^2}x} \right) - 2\cos x.\sqrt 3 \sin x}}{{2\cos 2x}} = \sqrt 3 \sin x - \cos x\\
\Leftrightarrow \dfrac{{{{\cos }^2}x - 2\cos x.\sqrt 3 \sin x + 3{{\sin }^2}x}}{{2\cos 2x}} = \sqrt 3 \sin x - \cos x\\
\Leftrightarrow \dfrac{{{{\left( {\cos x - \sqrt 3 \sin x} \right)}^2}}}{{2\cos 2x}} = \sqrt 3 \sin x - \cos x\\
\Leftrightarrow \left( {\cos x - \sqrt 3 \sin x} \right)\left( {\cos x - \sqrt 3 \sin x + 2\cos 2x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x - \sqrt 3 \sin x = 0\\
\cos x - \sqrt 3 \sin x + 2\cos 2x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{1}{2}\cos x - \dfrac{{\sqrt 3 }}{2}\sin x = 0\\
\cos 2x = \dfrac{1}{2}\cos x - \dfrac{{\sqrt 3 }}{2}\sin x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\cos \left( {x + \dfrac{\pi }{3}} \right) = 0\\
\cos 2x = \cos \left( {x + \dfrac{\pi }{3}} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{3} = \dfrac{\pi }{2} + k\pi \\
x + \dfrac{\pi }{3} = 2x + k2\pi \\
x + \dfrac{\pi }{3} = - 2x + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k\pi \\
x = \dfrac{\pi }{3} + k2\pi \\
x = - \dfrac{\pi }{9} + k\dfrac{{2\pi }}{3}
\end{array} \right. (tm)
\end{array}$
Vậy phương trình có các họ nghiệm là: ${x = \dfrac{\pi }{6} + k\pi \left( {k \in Z} \right);x = \dfrac{\pi }{3} + k2\pi \left( {k \in Z} \right);x = - \dfrac{\pi }{9} + k\dfrac{{2\pi }}{3}\left( {k \in Z} \right)}$
