Đáp án:
\[x = \dfrac{\pi }{6} + k2\pi \,\,\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\cos x = \sin \left( {\dfrac{\pi }{2} - x} \right)\\
{\sin ^2}x + {\cos ^2}x = 1\\
{\cos ^2}\left( {\dfrac{\pi }{3} + x} \right) + 4\cos \left( {\dfrac{\pi }{6} - x} \right) = 4\\
\Leftrightarrow {\sin ^2}\left[ {\dfrac{\pi }{2} - \left( {\dfrac{\pi }{3} + x} \right)} \right] + 4\cos \left( {\dfrac{\pi }{6} - x} \right) = 4\\
\Leftrightarrow {\sin ^2}\left( {\dfrac{\pi }{6} - x} \right) + 4\cos \left( {\dfrac{\pi }{6} - x} \right) = 4\\
\Leftrightarrow \left[ {1 - {{\cos }^2}\left( {\dfrac{\pi }{6} - x} \right)} \right] + 4\cos \left( {\dfrac{\pi }{6} - x} \right) = 4\\
\Leftrightarrow {\cos ^2}\left( {\dfrac{\pi }{6} - x} \right) - 4\cos \left( {\dfrac{\pi }{6} - x} \right) + 3 = 0\\
\Leftrightarrow \left( {\cos \left( {\dfrac{\pi }{6} - x} \right) - 1} \right)\left( {\cos \left( {\dfrac{\pi }{6} - x} \right) - 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos \left( {\dfrac{\pi }{6} - x} \right) = 1\\
\cos \left( {\dfrac{\pi }{6} - x} \right) = 3\,\,\,\,\,\,\,\,\,\,\,\left( {L,\,\,\, - 1 \le \cos x \le 1} \right)
\end{array} \right.\\
\Leftrightarrow \cos \left( {\dfrac{\pi }{6} - x} \right) = 1\\
\Leftrightarrow \dfrac{\pi }{6} - x = k2\pi \\
\Leftrightarrow x = \dfrac{\pi }{6} + k2\pi \,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
