If the function $f(x) = \dfrac{4\sqrt{2}(\sin 3x + \sin x)}{2\sin 2x\sin\dfrac{3x}{2} + \cos\dfrac{5x}{2} - \cos\dfrac{3x}{2}}$ for $x \neq \dfrac{\pi}{2}$ is continuous at $x = \dfrac{\pi}{2}$, then the value of $f\left(\dfrac{\pi}{2}\right)$ is equal to

Given the function:

$f(x) = \dfrac{4\sqrt{2}(\sin 3x + \sin x)}{2\sin 2x\sin\dfrac{3x}{2} + \cos\dfrac{5x}{2} - \cos\dfrac{3x}{2}}$

Using the sum-to-product formula $\sin C + \sin D = 2\sin\dfrac{C+D}{2}\cos\dfrac{C-D}{2}$, the numerator simplifies to:

$4\sqrt{2}(2\sin 2x \cos x) = 8\sqrt{2}\sin 2x \cos x$

For the denominator, using the formula $\cos C - \cos D = -2\sin\dfrac{C+D}{2}\sin\dfrac{C-D}{2}$, we get:

$\cos\dfrac{5x}{2} - \cos\dfrac{3x}{2} = -2\sin 2x \sin\dfrac{x}{2}$

Substituting this back into the denominator:

$2\sin 2x\sin\dfrac{3x}{2} - 2\sin 2x \sin\dfrac{x}{2} = 2\sin 2x \left(\sin\dfrac{3x}{2} - \sin\dfrac{x}{2}\right)$

Using the formula $\sin C - \sin D = 2\cos\dfrac{C+D}{2}\sin\dfrac{C-D}{2}$, the expression inside the parenthesis becomes:

$\sin\dfrac{3x}{2} - \sin\dfrac{x}{2} = 2\cos x \sin\dfrac{x}{2}$

So, the denominator simplifies to:

$2\sin 2x (2\cos x \sin\dfrac{x}{2}) = 4\sin 2x \cos x \sin\dfrac{x}{2}$

Now, substituting the simplified numerator and denominator back into $f(x)$:

$f(x) = \dfrac{8\sqrt{2}\sin 2x \cos x}{4\sin 2x \cos x \sin\dfrac{x}{2}}$

For $x \neq \dfrac{\pi}{2}$, we can cancel $\sin 2x \cos x$ from the numerator and denominator:

$f(x) = \dfrac{2\sqrt{2}}{\sin\dfrac{x}{2}}$

Since $f(x)$ is continuous at $x = \dfrac{\pi}{2}$, its value at this point is equal to its limit as $x \to \dfrac{\pi}{2}$:

$f\left(\dfrac{\pi}{2}\right) = \lim_{x \to \pi/2} \dfrac{2\sqrt{2}}{\sin\dfrac{x}{2}} = \dfrac{2\sqrt{2}}{\sin\dfrac{\pi}{4}} = \dfrac{2\sqrt{2}}{\dfrac{1}{\sqrt{2}}} = 4$

Therefore, $f\left(\dfrac{\pi}{2}\right) = 4 = (2)^2$.

Answer: $(2)^2$