$pt\Leftrightarrow \cos 7x+\sqrt3\sin 7x = \sqrt3\cos 5x +\sin 7x\Leftrightarrow \dfrac12\cos 7x+\dfrac{\sqrt3}2\sin 7x=\dfrac{\sqrt3}2\cos 5x+\dfrac12\sin 5x\Leftrightarrow \sin\left (7x+\dfrac{\pi}6\right) = \sin\left (5x+\dfrac{\pi}3\right)\\\Leftrightarrow
\left[\begin{matrix}
7x+\dfrac{\pi}6=5x+\dfrac{\pi}3+k2\pi\\
7x+\dfrac{\pi}6=\pi-5x-\dfrac{\pi}3+k2\pi
\end{matrix}\right.
\\\Leftrightarrow
\left[\begin{matrix}
2x=\dfrac{\pi}6+k2\pi\\
12x=\dfrac{\pi}2+k2\pi
\end{matrix}\right.
\\\Leftrightarrow
\left[\begin{matrix}
x=\dfrac{\pi}{12}+k\pi\\
x=\dfrac{\pi}{24}+\dfrac{k\pi}{6}
\end{matrix}\right.$
(với $k\in \mathbb{Z}$)