Solve `sin 3 theta-sin theta=4 cos^(2) theta-2`.

Explanation

To solve the equation \( \sin 3\theta - \sin \theta = 4 \cos^2 \theta - 2 \), we will follow a step-by-step approach.

### Step 1: Rewrite the left-hand side using the sine subtraction formula
We can use the sine subtraction formula:
\[ \sin A - \sin B = 2 \cos \left( \frac{A + B}{2} \right) \sin \left( \frac{A - B}{2} \right) \]
In our case, let \( A = 3\theta \) and \( B = \theta \):
\[ \sin 3\theta - \sin \theta = 2 \cos \left( \frac{3\theta + \theta}{2} \right) \sin \left( \frac{3\theta - \theta}{2} \right) \]
This simplifies to:
\[ \sin 3\theta - \sin \theta = 2 \cos (2\theta) \sin (\theta) \]

### Step 2: Rewrite the right-hand side
The right-hand side is:
\[ 4 \cos^2 \theta - 2 \]
We can factor this as:
\[ 2(2 \cos^2 \theta - 1) = 2 \cos(2\theta) \]

### Step 3: Set the two sides equal
Now we can set the two sides equal:
\[ 2 \cos(2\theta) \sin(\theta) = 2 \cos(2\theta) \]

### Step 4: Divide both sides by 2 (assuming \( \cos(2\theta) \neq 0 \))
Dividing both sides by 2 gives:
\[ \cos(2\theta) \sin(\theta) = \cos(2\theta) \]

### Step 5: Rearranging the equation
We can rearrange this to:
\[ \cos(2\theta) \sin(\theta) - \cos(2\theta) = 0 \]
Factoring out \( \cos(2\theta) \):
\[ \cos(2\theta)(\sin(\theta) - 1) = 0 \]

### Step 6: Solve for \( \theta \)
This gives us two cases to solve:

**Case 1:** \( \cos(2\theta) = 0 \)

The general solution for \( \cos(2\theta) = 0 \) is:
\[ 2\theta = \frac{(2n + 1)\pi}{2} \quad \text{for } n \in \mathbb{Z} \]
Thus,
\[ \theta = \frac{(2n + 1)\pi}{4} \quad \text{for } n \in \mathbb{Z} \]

**Case 2:** \( \sin(\theta) - 1 = 0 \)

This gives:
\[ \sin(\theta) = 1 \]
The general solution for \( \sin(\theta) = 1 \) is:
\[ \theta = \frac{\pi}{2} + 2k\pi \quad \text{for } k \in \mathbb{Z} \]

### Final Solution
Combining both cases, the solutions for \( \theta \) are:
\[ \theta = \frac{(2n + 1)\pi}{4} \quad \text{and} \quad \theta = \frac{\pi}{2} + 2k\pi \quad \text{for } n, k \in \mathbb{Z} \]