`sin^2 n theta- sin^2 (n-1)theta= sin^2 theta` where `n` is constant and `n != 0,1`

Explanation

To solve the equation \( \sin^2 n\theta - \sin^2 (n-1)\theta = \sin^2 \theta \), we can follow these steps:

### Step 1: Use the Difference of Squares Formula
We can rewrite the left-hand side using the difference of squares formula:
\[ a^2 - b^2 = (a - b)(a + b) \]
Let \( a = \sin n\theta \) and \( b = \sin (n-1)\theta \). Thus, we have:
\[ \sin^2 n\theta - \sin^2 (n-1)\theta = (\sin n\theta - \sin (n-1)\theta)(\sin n\theta + \sin (n-1)\theta) \]
So, the equation becomes:
\[ (\sin n\theta - \sin (n-1)\theta)(\sin n\theta + \sin (n-1)\theta) = \sin^2 \theta \]

### Step 2: Express the Difference of Sines
Using the sine difference identity, we can express \( \sin n\theta - \sin (n-1)\theta \):
\[ \sin n\theta - \sin (n-1)\theta = 2 \cos\left(\frac{n + (n-1)}{2}\theta\right) \sin\left(\frac{n - (n-1)}{2}\theta\right) = 2 \cos\left((n - \frac{1}{2})\theta\right) \sin\left(\frac{1}{2}\theta\right) \]

### Step 3: Substitute Back into the Equation
Substituting this back into our equation gives:
\[ 2 \cos\left((n - \frac{1}{2})\theta\right) \sin\left(\frac{1}{2}\theta\right)(\sin n\theta + \sin (n-1)\theta) = \sin^2 \theta \]

### Step 4: Analyze the Equation
Now we can analyze the equation:
1. \( \sin \theta = 0 \) gives \( \theta = k\pi \) for \( k \in \mathbb{Z} \).
2. The other part of the equation can be simplified to find \( \sin\left(\frac{1}{2}\theta\right) \) and \( \cos\left((n - \frac{1}{2})\theta\right) \).

### Step 5: Solve for \( \theta \)
From \( \sin\left(\frac{1}{2}\theta\right) = 0 \), we get:
\[ \frac{1}{2}\theta = m\pi \implies \theta = 2m\pi \quad (m \in \mathbb{Z}) \]
From \( \cos\left((n - \frac{1}{2})\theta\right) = 0 \), we have:
\[ (n - \frac{1}{2})\theta = \frac{\pi}{2} + q\pi \implies \theta = \frac{(2q + 1)\pi}{2(n - \frac{1}{2}} \quad (q \in \mathbb{Z}) \]

### Final Result
Thus, the solutions for \( \theta \) are:
1. \( \theta = k\pi \) for \( k \in \mathbb{Z} \)
2. \( \theta = \frac{(2q + 1)\pi}{2(n - \frac{1}{2})} \) for \( q \in \mathbb{Z} \)