$\begin{array}{l}
151)\cos \left( {\frac{\pi }{2} + 2x} \right) - \sqrt 3 \cos \left( {\pi - 2x} \right) = 1\\
\Leftrightarrow - \sin 2x + \sqrt 3 \cos 2x = 1 \Leftrightarrow \frac{{\sqrt 3 }}{2}\cos 2x - \frac{1}{2}\sin 2x = \frac{1}{2}\\
\Leftrightarrow \cos \left( {2x + \frac{\pi }{6}} \right) = \frac{1}{2} \Leftrightarrow \left[ \begin{array}{l}
2x + \frac{\pi }{6} = \frac{\pi }{3} + k2\pi \\
2x + \frac{\pi }{6} = - \frac{\pi }{3} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{{12}} + k\pi \\
x = - \frac{\pi }{4} + k\pi
\end{array} \right.
\end{array}$
\[\begin{array}{l}
152)\\
\cos 2x + \sin x = \sqrt 3 \left( {\cos x - \sin 2x} \right)\\
\Leftrightarrow \cos 2x + \sqrt 3 \sin 2x = \sqrt 3 \cos x - \sin x\\
\Leftrightarrow \frac{1}{2}\cos 2x + \frac{{\sqrt 3 }}{2}\sin 2x = \frac{{\sqrt 3 }}{2}\cos x - \frac{1}{2}\sin x\\
\Leftrightarrow \cos \left( {2x - \frac{\pi }{3}} \right) = \cos \left( {x + \frac{\pi }{6}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
2x - \frac{\pi }{3} = x + \frac{\pi }{6} + k2\pi \\
2x - \frac{\pi }{3} = - x - \frac{\pi }{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{2} + k2\pi \\
x = \frac{\pi }{{18}} + \frac{{k2\pi }}{3}
\end{array} \right.
\end{array}\]
153)
\[\begin{array}{l}
2\left( {\cos x + \sqrt 3 \sin x} \right)\cos x = \cos x - \sqrt 3 \sin x + 1\\
\Leftrightarrow 2{\cos ^2}x - 1 + 2\sqrt 3 \sin x\cos x = \cos x - \sqrt 3 \sin x\\
\Leftrightarrow \cos 2x + \sqrt 3 \sin 2x = \cos x - \sqrt 3 \sin x\\
\Leftrightarrow \frac{1}{2}\cos 2x + \frac{{\sqrt 3 }}{2}\sin 2x = \frac{1}{2}\cos x - \frac{{\sqrt 3 }}{2}\sin x\\
\Leftrightarrow \cos \left( {2x - \frac{\pi }{3}} \right) = \cos \left( {x + \frac{\pi }{6}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
2x - \frac{\pi }{3} = x + \frac{\pi }{6} + k2\pi \\
2x - \frac{\pi }{3} = - x - \frac{\pi }{6} + k2\pi
\end{array} \right.
\end{array}\]
Bạn làm tiếp nhé!
