Let $$S = \left\{\theta \in [-\pi, \pi] : \cos\theta \cos\left(\frac{5\theta}{2}\right) = \cos 7\theta \cos\left(\frac{7\theta}{2}\right)\right\}$$. Then $$n(S)$$ is equal to :

We have to solve the equation

$$\cos\theta \,\cos\!\left(\frac{5\theta}{2}\right)=\cos 7\theta \,\cos\!\left(\frac{7\theta}{2}\right),\qquad \theta\in[-\pi,\pi].$$

Use the product-to-sum identity $$\cos A\cos B=\tfrac12\,[\cos(A+B)+\cos(A-B)].$$

Left side:
$$\cos\theta\cos\!\left(\frac{5\theta}{2}\right)=\tfrac12\!\left[\cos\!\left(\theta+\frac{5\theta}{2}\right)+\cos\!\left(\theta-\frac{5\theta}{2}\right)\right]$$ $$=\tfrac12\,[\cos(7\theta/2)+\cos(3\theta/2)].$$

Right side:
$$\cos7\theta\cos\!\left(\frac{7\theta}{2}\right)=\tfrac12\!\left[\cos\!\left(7\theta+\frac{7\theta}{2}\right)+\cos\!\left(7\theta-\frac{7\theta}{2}\right)\right]$$ $$=\tfrac12\,[\cos(21\theta/2)+\cos(7\theta/2)].$$

Equating and multiplying by 2 gives

$$\cos\!\left(\frac{7\theta}{2}\right)+\cos\!\left(\frac{3\theta}{2}\right)= \cos\!\left(\frac{21\theta}{2}\right)+\cos\!\left(\frac{7\theta}{2}\right).$$

The terms $$\cos(7\theta/2)$$ cancel, leaving

$$\cos\!\left(\frac{3\theta}{2}\right)=\cos\!\left(\frac{21\theta}{2}\right).$$

For $$\cos\alpha=\cos\beta$$ the general solutions are $$\alpha=2k\pi\pm\beta,\;k\in\mathbb Z.$$
Put $$\alpha=\frac{3\theta}{2},\;\beta=\frac{21\theta}{2}:$$

$$\frac{3\theta}{2}=2k\pi+\frac{21\theta}{2}\;\Longrightarrow\; \frac{3\theta}{2}-\frac{21\theta}{2}=2k\pi\;\Longrightarrow\; -9\theta=2k\pi\;\Longrightarrow\; \theta=-\frac{2k\pi}{9}.$$

$$\frac{3\theta}{2}=2k\pi-\frac{21\theta}{2}\;\Longrightarrow\; \frac{3\theta}{2}+\frac{21\theta}{2}=2k\pi\;\Longrightarrow\; 12\theta=2k\pi\;\Longrightarrow\; \theta=\frac{k\pi}{6}.$$

Thus the solution set is

$$S=\Bigl\{\theta=-\tfrac{2k\pi}{9}\Bigr\}\cup \Bigl\{\theta=\tfrac{k\pi}{6}\Bigr\},\qquad k\in\mathbb Z,$$ subject to $$\theta\in[-\pi,\pi].$$

Counting solutions from Case 1 (family A):
$$|\theta|=\frac{2\pi|k|}{9}\le\pi\; \Longrightarrow\; |k|\le\frac{9}{2}.$$
Hence $$k=-4,-3,-2,-1,0,1,2,3,4$$  ⇒  9 values.

Counting solutions from Case 2 (family B):
$$-\pi\le\frac{k\pi}{6}\le\pi\; \Longrightarrow\; -6\le k\le 6.$$
Hence $$k=-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6$$  ⇒  13 values.

Removing duplicates
Common angles in both families are $$\theta=0,\;\theta=\pm\frac{2\pi}{3},$$ that is 3 overlaps.

Therefore

$$n(S)=13+9-3=19.$$

Hence the required number of solutions is 19.