Solve the general vlaue.
`3-2cos theta -4sin theta - cos 2 theta + sin 2 theta =0`

Answer

To solve the equation \(3 - 2\cos \theta - 4\sin \theta - \cos 2\theta + \sin 2\theta = 0\), we will simplify and rearrange the terms step by step.

### Step 1: Rewrite the trigonometric identities
We start by rewriting \(\cos 2\theta\) and \(\sin 2\theta\) using trigonometric identities:
\[ \cos 2\theta = 1 - 2\sin^2 \theta \]
\[ \sin 2\theta = 2\sin \theta \cos \theta \]
Substituting these 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, we simplify the equation:
\[ 3 - 2\cos \theta - 4\sin \theta - 1 + 2\sin^2 \theta + 2\sin \theta \cos \theta = 0 \]
Combining like terms:
\[ 2 - 2\cos \theta - 4\sin \theta + 2\sin^2 \theta + 2\sin \theta \cos \theta = 0 \]

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

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

### Step 5: Grouping terms
We can group the terms involving \(\sin \theta\):
\[ \sin^2 \theta + \sin \theta (\cos \theta - 2) - \cos \theta + 1 = 0 \]

### Step 6: Solve for \(\sin \theta\)
Now, we can treat this as a quadratic in \(\sin \theta\):
Let \(x = \sin \theta\), then we have:
\[ x^2 + x(\cos \theta - 2) - \cos \theta + 1 = 0 \]
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
Here, \(a = 1\), \(b = \cos \theta - 2\), and \(c = -\cos \theta + 1\).

### Step 7: Find the roots
Calculating the discriminant:
\[ D = (\cos \theta - 2)^2 - 4(1)(-\cos \theta + 1) \]
\[ D = \cos^2 \theta - 4\cos \theta + 4 + 4\cos \theta - 4 = \cos^2 \theta \]
Thus, the roots are:
\[ x = \frac{-(\cos \theta - 2) \pm \sqrt{\cos^2 \theta}}{2} \]
This simplifies to:
\[ x = \frac{2 - \cos \theta \pm \cos \theta}{2} \]
Thus, we have two cases:
1. \(x = 1\) (which gives \(\sin \theta = 1\))
2. \(x = 2 - \cos \theta\)

### Step 8: Solve for \(\theta\)
For \(\sin \theta = 1\):
\[ \theta = \frac{\pi}{2} + 2n\pi, \quad n \in \mathbb{Z} \]

For \(2 - \cos \theta = \sin \theta\):
This can be solved using the identity \(\sin^2 \theta + \cos^2 \theta = 1\).

### Final Solution
The general solutions for the given equation are:
\[ \theta = \frac{\pi}{2} + 2n\pi \quad \text{and} \quad \sin \theta + \cos \theta = 1 \]