Number of Solutions to Trigonometric Equation in Given Interval

To find the number of elements in the set $S = \left\{ \theta \in [-\pi, \pi] : \cos\theta \cos \frac{5\theta}{2} = \cos 7\theta \cos \frac{7\theta}{2} \right\}$, we start by simplifying the given trigonometric equation.

Given equation: $\cos\theta \cos \frac{5\theta}{2} = \cos 7\theta \cos \frac{7\theta}{2}$

Multiplying both sides by $2$: $2\cos\theta \cos \frac{5\theta}{2} = 2\cos 7\theta \cos \frac{7\theta}{2}$

Using the product-to-sum identity $2\cos A \cos B = \cos(A+B) + \cos(A-B)$:

For the Left-Hand Side (LHS): $2\cos\theta \cos \frac{5\theta}{2} = \cos\left(\frac{5\theta}{2} + \theta\right) + \cos\left(\frac{5\theta}{2} - \theta\right) = \cos\frac{7\theta}{2} + \cos\frac{3\theta}{2}$

For the Right-Hand Side (RHS): $2\cos 7\theta \cos \frac{7\theta}{2} = \cos\left(7\theta + \frac{7\theta}{2}\right) + \cos\left(7\theta - \frac{7\theta}{2}\right) = \cos\frac{21\theta}{2} + \cos\frac{7\theta}{2}$

Equating LHS and RHS: $\cos\frac{7\theta}{2} + \cos\frac{3\theta}{2} = \cos\frac{21\theta}{2} + \cos\frac{7\theta}{2}$

Subtracting $\cos\frac{7\theta}{2}$ from both sides gives: $\cos\frac{3\theta}{2} = \cos\frac{21\theta}{2}$

Rearranging the terms: $\cos\frac{21\theta}{2} - \cos\frac{3\theta}{2} = 0$

Using the sum-to-product identity $\cos C - \cos D = -2\sin\left(\frac{C+D}{2}\right)\sin\left(\frac{C-D}{2}\right)$: $-2\sin\left(\frac{\frac{21\theta}{2} + \frac{3\theta}{2}}{2}\right)\sin\left(\frac{\frac{21\theta}{2} - \frac{3\theta}{2}}{2}\right) = 0$ $-2\sin(6\theta)\sin\left(\frac{9\theta}{2}\right) = 0$

Thus, the equation holds if either $\sin(6\theta) = 0$ or $\sin\left(\frac{9\theta}{2}\right) = 0$.

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Case 1: $\sin(6\theta) = 0$

$6\theta = k\pi \implies \theta = \frac{k\pi}{6}, \quad \text{where } k \in \mathbb{Z}$

Since $\theta \in [-\pi, \pi]$: $-\pi \le \frac{k\pi}{6} \le \pi \implies -6 \le k \le 6$ The set of integer values for $k$ is: $k \in \{-6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6\}$ This gives $13$ solutions.

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Case 2: $\sin\left(\frac{9\theta}{2}\right) = 0$

$\frac{9\theta}{2} = m\pi \implies \theta = \frac{2m\pi}{9}, \quad \text{where } m \in \mathbb{Z}$

Since $\theta \in [-\pi, \pi]$: $-\pi \le \frac{2m\pi}{9} \le \pi \implies -\frac{9}{2} \le m \le \frac{9}{2}$ The set of integer values for $m$ is: $m \in \{-4, -3, -2, -1, 0, 1, 2, 3, 4\}$ This gives $9$ solutions.

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Finding the Common Solutions (Intersection):

To avoid double-counting, we find the common solutions where: $\frac{k\pi}{6} = \frac{2m\pi}{9} \implies 3k = 4m$

Since $\gcd(3, 4) = 1$, $k$ must be a multiple of $4$. Therefore, $k = 4p$ for some $p \in \mathbb{Z}$. Substituting $k = 4p$ gives $m = 3p$.

For $k \in [-6, 6]$ and $m \in [-4, 4]$:

• For $p = 0 \implies k = 0, m = 0 \implies \theta = 0$
• For $p = 1 \implies k = 4, m = 3 \implies \theta = \frac{2\pi}{3}$
• For $p = -1 \implies k = -4, m = -3 \implies \theta = -\frac{2\pi}{3}$

Other values of $p$ place $k$ and $m$ outside their allowable ranges. Thus, there are $3$ common solutions.

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Conclusion:

Using the Principle of Inclusion-Exclusion, the total number of distinct solutions $n(S)$ is: $n(S) = 13 + 9 - 3 = 19$