Đáp án:
\[\left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + k\pi \\
x = - \dfrac{\pi }{4} + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\cos x = \sin \left( {\dfrac{\pi }{2} - x} \right)\\
\cos x = - \cos \left( {\pi - x} \right)\\
\sin x = - \sin \left( { - x} \right)\\
\cos \left( {\dfrac{\pi }{2} + 2x} \right) - \sqrt 3 \cos \left( {\pi - 2x} \right) = 1\\
\Leftrightarrow \sin \left[ {\dfrac{\pi }{2} - \left( {\dfrac{\pi }{2} + 2x} \right)} \right] - \sqrt 3 .\left( { - \cos 2x} \right) = 1\\
\Leftrightarrow \sin \left( { - 2x} \right) + \sqrt 3 \cos 2x = 1\\
\Leftrightarrow - \sin 2x + \sqrt 3 \cos 2x = 1\\
\Leftrightarrow \sqrt 3 \cos 2x - \sin 2x = 1\\
\Leftrightarrow \dfrac{{\sqrt 3 }}{2}\cos 2x - \dfrac{1}{2}\sin 2x = \dfrac{1}{2}\\
\Leftrightarrow \cos 2x.\cos \dfrac{\pi }{6} - \sin 2x.\sin \dfrac{\pi }{6} = \dfrac{1}{2}\\
\Leftrightarrow \cos \left( {2x + \dfrac{\pi }{6}} \right) = \cos \dfrac{\pi }{3}\\
\Leftrightarrow \left[ \begin{array}{l}
2x + \dfrac{\pi }{6} = \dfrac{\pi }{3} + k2\pi \\
2x + \dfrac{\pi }{6} = - \dfrac{\pi }{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{6} + k2\pi \\
2x = - \dfrac{\pi }{2} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + k\pi \\
x = - \dfrac{\pi }{4} + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
