Determine Positive Integer Value Satisfying Trigonometric Reciprocal Equation

To find the positive integer value of $n > 3$ satisfying the given equation:
$\frac{1}{\sin\left(\frac{\pi}{n}\right)} = \frac{1}{\sin\left(\frac{2\pi}{n}\right)} + \frac{1}{\sin\left(\frac{3\pi}{n}\right)}$

Let $\theta = \frac{\pi}{n}$. For $n > 3$, we have $0 < \theta < \frac{\pi}{3}$, which ensures that $\sin\theta, \sin 2\theta, \sin 3\theta \neq 0$.

Rewriting the equation in terms of $\theta$:
$\frac{1}{\sin\theta} - \frac{1}{\sin 3\theta} = \frac{1}{\sin 2\theta}$

Combining the terms on the left-hand side:
$\frac{\sin 3\theta - \sin\theta}{\sin\theta \sin 3\theta} = \frac{1}{\sin 2\theta}$

Using the sum-to-product formula $\sin A - \sin B = 2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$:
$\frac{2\cos 2\theta \sin\theta}{\sin\theta \sin 3\theta} = \frac{1}{\sin 2\theta}$

Since $\sin\theta \neq 0$, we can cancel $\sin\theta$ from the numerator and denominator:
$\frac{2\cos 2\theta}{\sin 3\theta} = \frac{1}{\sin 2\theta}$

Cross-multiplying gives:
$2\sin 2\theta \cos 2\theta = \sin 3\theta$

Using the double-angle formula $\sin 2A = 2\sin A \cos A$:
$\sin 4\theta = \sin 3\theta$

Rearranging the terms:
$\sin 4\theta - \sin 3\theta = 0$
$2\cos\left(\frac{7\theta}{2}\right)\sin\left(\frac{\theta}{2}\right) = 0$

Since $0 < \theta < \frac{\pi}{3}$, we have $0 < \frac{\theta}{2} < \frac{\pi}{6}$, which implies $\sin\left(\frac{\theta}{2}\right) \neq 0$.

Therefore, we must have:
$\cos\left(\frac{7\theta}{2}\right) = 0$

The general solution is:
$\frac{7\theta}{2} = (2k + 1)\frac{\pi}{2}, \quad k \in \mathbb{Z}$
$7\theta = (2k + 1)\pi$

Substituting $\theta = \frac{\pi}{n}$:
$\frac{7\pi}{n} = (2k + 1)\pi \implies n = \frac{7}{2k + 1}$

For $n$ to be an integer greater than $3$, the only possible non-negative integer value for $k$ is $k = 0$:
$n = \frac{7}{2(0) + 1} = 7$

Thus, the required positive integer value of $n$ is $7$.