Solve : `3-2 cos theta -4 sin theta - cos 2theta+sin 2theta=0`.

Explanation

To solve the equation \( 3 - 2 \cos \theta - 4 \sin \theta - \cos 2\theta + \sin 2\theta = 0 \), we can follow these steps:

### Step 1: Rewrite the equation using trigonometric identities
We know that:
- \( \cos 2\theta = 1 - 2\sin^2 \theta \)
- \( \sin 2\theta = 2 \sin \theta \cos \theta \)

Substituting these identities into the equation gives:
\[ 3 - 2 \cos \theta - 4 \sin \theta - (1 - 2 \sin^2 \theta) + 2 \sin \theta \cos \theta = 0 \]

### Step 2: Simplify the equation
Now, simplify the equation:
\[ 3 - 2 \cos \theta - 4 \sin \theta - 1 + 2 \sin^2 \theta + 2 \sin \theta \cos \theta = 0 \]
This simplifies to:
\[ 2 + 2 \sin^2 \theta - 2 \cos \theta - 4 \sin \theta + 2 \sin \theta \cos \theta = 0 \]

### Step 3: Factor out common terms
We can factor out 2 from the entire equation:
\[ 2(\sin^2 \theta - 2 \sin \theta + \sin \theta \cos \theta - \cos \theta + 1) = 0 \]
This implies:
\[ \sin^2 \theta - 2 \sin \theta + \sin \theta \cos \theta - \cos \theta + 1 = 0 \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ \sin^2 \theta + \sin \theta \cos \theta - 2 \sin \theta - \cos \theta + 1 = 0 \]

### Step 5: Solve for \(\sin \theta\) and \(\cos \theta\)
We can treat this as a quadratic in \(\sin \theta\):
Let \(x = \sin \theta\), then we can rewrite the equation as:
\[ x^2 + x \cos \theta - 2x - \cos \theta + 1 = 0 \]

### Step 6: Use the quadratic formula
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = \cos \theta - 2\), and \(c = -\cos \theta + 1\):
\[ x = \frac{-(\cos \theta - 2) \pm \sqrt{(\cos \theta - 2)^2 - 4(1)(-\cos \theta + 1)}}{2(1)} \]

### Step 7: Solve for \(\theta\)
Now we can solve for \(\theta\) using the values of \(x\) obtained from the quadratic equation.

### Step 8: Finding specific solutions
1. Set \( \sin \theta = 1 \):
\[ \theta = \frac{\pi}{2} + 2n\pi \quad (n \in \mathbb{Z}) \]

2. Set \( \sin \theta + \cos \theta = 1 \):
\[ \sin \theta = 1 - \cos \theta \]
Substitute into the Pythagorean identity:
\[ (1 - \cos \theta)^2 + \cos^2 \theta = 1 \]
This leads to further simplifications and solutions.

### Final Solutions
Thus, the solutions for \(\theta\) are:
\[ \theta = \frac{\pi}{2} + 2n\pi \quad \text{and other solutions from the quadratic.} \]