The number of solutions of the equation $$1 + \sin^4 x = \cos^2 3x$$, $$x \in \left[-\frac{5\pi}{2}, \frac{5\pi}{2}\right]$$ is:

We have to solve the trigonometric equation

$$1+\sin^{4}x=\cos^{2}3x$$

for all real $$x$$ lying in the closed interval $$\left[-\dfrac{5\pi}{2},\dfrac{5\pi}{2}\right]$$.

First, note the possible ranges of the two sides.

For any real number $$y$$, the basic facts are

$$0\le\sin^{2}y\le1\quad\text{and hence}\quad0\le\sin^{4}y\le1.$$

Therefore

$$1+\sin^{4}x\in[1,2].$$

Similarly, for any real number $$z$$, we know

$$-1\le\cos z\le1\quad\Longrightarrow\quad0\le\cos^{2}z\le1.$$

So the right‐hand side satisfies

$$\cos^{2}3x\in[0,1].$$

Now we compare the two ranges. The left‐hand side is always at least $$1$$, while the right‐hand side is at most $$1$$. Hence equality is possible only when both quantities are exactly $$1$$. Therefore we must have

$$1+\sin^{4}x=1\quad\text{and}\quad\cos^{2}3x=1.$$

The first condition gives

$$\sin^{4}x=0\quad\Longrightarrow\quad\sin x=0.$$

The standard solutions of $$\sin x=0$$ are

$$x=n\pi,\qquad n\in\mathbb Z.$$

Next, we verify the second condition at these points. Substituting $$x=n\pi$$ in $$\cos^{2}3x$$, we get

$$\cos^{2}3x=\cos^{2}(3n\pi)=\bigl(\cos(3n\pi)\bigr)^{2}.$$

But $$\cos(3n\pi)=(-1)^{3n}=(-1)^{n},$$ and squaring eliminates the sign, giving

$$\cos^{2}(3n\pi)=\bigl((-1)^{n}\bigr)^{2}=1.$$

Thus every $$x=n\pi$$ automatically satisfies $$\cos^{2}3x=1$$, so all such points are indeed solutions.

We now count how many integers $$n$$ give $$x=n\pi$$ in the required interval. Divide the inequality

$$-\dfrac{5\pi}{2}\le n\pi\le\dfrac{5\pi}{2}$$

by $$\pi$$ (which is positive, so the inequality sign does not change):

$$-\dfrac{5}{2}\le n\le\dfrac{5}{2}.$$

The integers in the interval $$[-2.5,\,2.5]$$ are

$$n=-2,\,-1,\,0,\,1,\,2.$$

Counting them gives exactly $$5$$ values of $$n$$, hence $$5$$ solutions $$x=n\pi$$ in the given domain.

Hence, the correct answer is Option A.