1. Given equation

We need to solve

$$\cos 2\theta \cos \frac{\theta}{2}+\cos \frac{5\theta}{2}=2\cos^3\frac{5\theta}{2}$$

in

$$\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right].$$

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2. Use the identity

Recall:

$$2\cos^3 x=\cos x+\cos x\cos 2x$$

because

$$2\cos^3 x=\cos x(1+2\cos^2 x-1)=\cos x+\cos x(2\cos^2 x-1)=\cos x+\cos x\cos 2x.$$

Putting $x=\frac{5\theta}{2}$,

$$2\cos^3\frac{5\theta}{2}=\cos\frac{5\theta}{2}+\cos\frac{5\theta}{2}\cos 5\theta.$$

So the equation becomes

$$\cos 2\theta\cos\frac{\theta}{2}+\cos\frac{5\theta}{2} =\cos\frac{5\theta}{2}+\cos\frac{5\theta}{2}\cos 5\theta.$$

Cancel $\cos\frac{5\theta}{2}$ from both sides:

$$\cos 2\theta\cos\frac{\theta}{2}=\cos\frac{5\theta}{2}\cos 5\theta.$$

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3. Expand product-to-sum

Use

$$\cos A\cos B=\frac{1}{2}[\cos(A+B)+\cos(A-B)].$$

Then

$$\cos 2\theta\cos\frac{\theta}{2} =\frac{1}{2}\left(\cos\frac{5\theta}{2}+\cos\frac{3\theta}{2}\right),$$

and

$$\cos\frac{5\theta}{2}\cos 5\theta =\frac{1}{2}\left(\cos\frac{15\theta}{2}+\cos\frac{5\theta}{2}\right).$$

Hence

\frac{1}{2}\left(\cos\frac{5\theta}{2}+\cos\frac{3\theta}{2}\right) =rac{1}{2}\left(\cos\frac{15\theta}{2}+\cos\frac{5\theta}{2}\right).

Multiply by $2$ and cancel $\cos\frac{5\theta}{2}$:

$$\cos\frac{3\theta}{2}=\cos\frac{15\theta}{2}.$$

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4. Solve $\cos A=\cos B$

If $\cos A=\cos B$, then

$$A=2n\pi\pm B,\qquad n\in\mathbb Z.$$

Here

$$A=\frac{3\theta}{2},\qquad B=\frac{15\theta}{2}.$$

So we get two cases.

Case 1:

$$\frac{3\theta}{2}=2n\pi+\frac{15\theta}{2}$$ $$-6\theta=2n\pi$$ $$\theta=-\frac{n\pi}{3}.$$

Case 2:

$$\frac{3\theta}{2}=2n\pi-\frac{15\theta}{2}$$ $$9\theta=2n\pi$$ $$\theta=\frac{2n\pi}{9}.$$

Thus all solutions are from

$$\theta=\frac{m\pi}{3}$$

or

$$\theta=\frac{2n\pi}{9},$$

with values restricted to $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$.

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5. Count solutions in the interval

From $\theta=-\frac{n\pi}{3}$ i.e. $\theta=\frac{m\pi}{3}$

Within

$$\left[-\frac{\pi}{2},\frac{\pi}{2}\right],$$

possible values are

$$-\frac{\pi}{3},\ 0,\ \frac{\pi}{3}.$$

So this gives 3 solutions.

From $\theta=\frac{2n\pi}{9}$

Need

$$-\frac{\pi}{2}\le \frac{2n\pi}{9}\le \frac{\pi}{2}.$$

So

$$-\frac{1}{2}\le \frac{2n}{9}\le \frac{1}{2}$$ $$-\frac{9}{4}\le n\le \frac{9}{4}.$$

Hence

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

This gives

$$-\frac{4\pi}{9},\ -\frac{2\pi}{9},\ 0,\ \frac{2\pi}{9},\ \frac{4\pi}{9}.$$

So this gives 5 solutions.

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6. Remove duplicates

The common value in the two sets is only

$$\theta=0.$$

So total number of distinct solutions is

$$3+5-1=7.$$

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7. Check with options

Thus the number of solutions is

$$\boxed{7}.$$

So the correct option is B.