Given $\cos 3x=\sin 4x$

$\Rightarrow \cos 3x=\cos \left(\frac{\mathrm{\pi }}{2}-4x\right)$

If $\cos \theta =\cos \alpha ,$ then $\theta =2n\pi \pm \alpha , n\in Z,$

$\Rightarrow 3x=2n\mathrm{\pi }\pm \left(\frac{\mathrm{\pi }}{2}-4\mathrm{x}\right)$

$\Rightarrow 3x=2n\mathrm{\pi }+\left(\frac{\mathrm{\pi }}{2}-4\mathrm{x}\right)$ or $3x=2n\mathrm{\pi }-\left(\frac{\mathrm{\pi }}{2}-4\mathrm{x}\right)$

$\Rightarrow 7x=2n\mathrm{\pi }+\frac{\mathrm{\pi }}{2}$ or $-x=2n\mathrm{\pi }-\frac{\mathrm{\pi }}{2}$

$\Rightarrow x=\frac{2\mathrm{n\pi }}{7}+\frac{\mathrm{\pi }}{14}$ or $x=-2n\mathrm{\pi }+\frac{\mathrm{\pi }}{2}$

Since, $-\frac{\mathrm{\pi }}{6}<x<\frac{\mathrm{\pi }}{2},$ hence, we get $x=\frac{\mathrm{\pi }}{14}, \frac{5\mathrm{\pi }}{14}$ by putting $n=0, 1$ in the first value and from the second value none of the value lie in the given interval.

Thus, there are total two solutions of the equation in the given interval.