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

The given equation is
$$\cos 2\theta \,\cos\frac{\theta}{2}+\cos\frac{5\theta}{2}=2\cos^{3}\frac{5\theta}{2},\qquad \theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right].$$

Put $$t=\frac{\theta}{2}.$$
Then $$\theta=2t$$ and the interval becomes $$t\in\left[-\frac{\pi}{4},\frac{\pi}{4}\right].$$
The equation turns into
$$\cos 4t\;\cos t+\cos 5t=2\cos^{3}5t \quad -(1).$$

Use the product-to-sum identity $$\cos A\cos B=\tfrac{1}{2}\bigl[\cos(A+B)+\cos(A-B)\bigr]$$ on the first term:
$$\cos 4t\;\cos t=\tfrac{1}{2}\bigl[\cos(4t+t)+\cos(4t-t)\bigr] =\tfrac{1}{2}\bigl[\cos 5t+\cos 3t\bigr].$$

Substitute this in $$(1):$$
$$\tfrac{1}{2}\bigl[\cos 5t+\cos 3t\bigr]+\cos 5t =2\cos^{3}5t.$$ Combine like terms:
$$(\tfrac{1}{2}+1)\cos 5t+\tfrac{1}{2}\cos 3t =2\cos^{3}5t,$$ $$\tfrac{3}{2}\cos 5t+\tfrac{1}{2}\cos 3t =2\cos^{3}5t.$$ Multiply by $$2:$$ $$3\cos 5t+\cos 3t=4\cos^{3}5t \quad -(2).$$

Recall the triple-angle identity $$\cos 3x=4\cos^{3}x-3\cos x,$$ which can be rearranged as
$$4\cos^{3}x=\cos 3x+3\cos x.$$ Putting $$x=5t$$ gives
$$4\cos^{3}5t=\cos 15t+3\cos 5t.$$ Replace the right side of $$(2)$$ with this expression:
$$3\cos 5t+\cos 3t=\cos 15t+3\cos 5t.$$ The terms $$3\cos 5t$$ cancel, leaving the simple equation
$$\cos 3t=\cos 15t \quad -(3).$$

For $$\cos\alpha=\cos\beta,$$ the solutions are
$$\alpha=2k\pi\pm\beta,\qquad k\in\mathbb{Z}.$$ Apply this to $$(3)$$:

$$3t=2k\pi+15t \;\Longrightarrow\; -12t=2k\pi \;\Longrightarrow\; t=-\frac{k\pi}{6}.$$ With $$t\in\left[-\frac{\pi}{4},\frac{\pi}{4}\right],$$ choose integers $$k$$ so that $$-\frac{k\pi}{6}$$ lies in the interval.
  • $$k=0:\;t=0$$
  • $$k=1:\;t=-\frac{\pi}{6}$$
  • $$k=-1:\;t=\frac{\pi}{6}$$
Further $$|k|\ge 2$$ gives $$|t|\gt\pi/4,$$ so they are rejected.
Thus Case 1 contributes $$t=-\frac{\pi}{6},\,0,\,\frac{\pi}{6}.$$

$$3t=2k\pi-15t \;\Longrightarrow\;18t=2k\pi \;\Longrightarrow\; t=\frac{k\pi}{9}.$$ Again keep the values inside $$\left[-\frac{\pi}{4},\frac{\pi}{4}\right]:$$
  • $$k=0:\;t=0$$ (already counted)
  • $$k=\pm1:\;t=\pm\frac{\pi}{9}$$
  • $$k=\pm2:\;t=\pm\frac{2\pi}{9}$$
  • $$|k|\ge3$$ violates the interval.
Case 2 contributes $$t=-\frac{2\pi}{9},\,-\frac{\pi}{9},\,0,\,\frac{\pi}{9},\,\frac{2\pi}{9}.$$

Collecting distinct solutions of $$t$$ inside the interval:
$$t=-\frac{2\pi}{9},\;-\frac{\pi}{6},\;-\frac{\pi}{9},\;0,\;\frac{\pi}{9},\;\frac{\pi}{6},\;\frac{2\pi}{9}.$$ Thus there are $$7$$ valid $$t$$ values.

Since $$\theta=2t,$$ each $$t$$ gives a unique $$\theta$$ in the original interval $$\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$$, so the number of solutions for $$\theta$$ is also $$7$$.

Hence, the correct option is Option A (7).