The number of solutions in $\left[0, \dfrac{\pi}{2}\right)$ of the equation $\cos 3x \cdot \tan 5x = \sin 7x$ is

Given equation: $\cos 3x \cdot \tan 5x = \sin 7x$

$\cos 3x \cdot \dfrac{\sin 5x}{\cos 5x} = \sin 7x$

$2 \sin 5x \cos 3x = 2 \sin 7x \cos 5x$

Using the identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$:

$\sin 8x + \sin 2x = \sin 12x + \sin 2x$

$\sin 12x - \sin 8x = 0$

Using the identity $\sin C - \sin D = 2 \sin\left(\dfrac{C-D}{2}\right) \cos\left(\dfrac{C+D}{2}\right)$:

$2 \sin 2x \cos 10x = 0$

This gives $\sin 2x = 0$ or $\cos 10x = 0$.

For $\sin 2x = 0$, $2x = n\pi \Rightarrow x = \dfrac{n\pi}{2}$.

Since $x \in \left[0, \dfrac{\pi}{2}\right)$, the only solution is $x = 0$.

For $\cos 10x = 0$, $10x = (2k+1)\dfrac{\pi}{2} \Rightarrow x = (2k+1)\dfrac{\pi}{20}$.

Since $x \in \left[0, \dfrac{\pi}{2}\right)$, we have $0 \leq (2k+1)\dfrac{\pi}{20} < \dfrac{\pi}{2} \Rightarrow 0 \leq 2k+1 < 10$.

The possible values for $2k+1$ are $1, 3, 5, 7, 9$.

This gives $x = \dfrac{\pi}{20}, \dfrac{3\pi}{20}, \dfrac{5\pi}{20}, \dfrac{7\pi}{20}, \dfrac{9\pi}{20}$.

We must also ensure that $\tan 5x$ is defined, which means $\cos 5x \neq 0$.

$\cos 5x = 0 \Rightarrow 5x = (2m+1)\dfrac{\pi}{2} \Rightarrow x = (4m+2)\dfrac{\pi}{20}$.

Since $2k+1$ is always odd and $4m+2$ is always even, none of our solutions make $\cos 5x = 0$.

Thus, all $1 + 5 = 6$ solutions are valid.

Answer: $6$