Solve `3 cos 2 theta-sin theta=2`

Answer

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

### Step 1: Rewrite \(\cos 2\theta\)
We know that \(\cos 2\theta\) can be expressed in terms of \(\sin \theta\):
\[ \cos 2\theta = 1 - 2\sin^2 \theta \]
Substituting this into the equation gives:
\[ 3(1 - 2\sin^2 \theta) - \sin \theta = 2 \]

### Step 2: Simplify the equation
Now, expand and simplify the equation:
\[ 3 - 6\sin^2 \theta - \sin \theta = 2 \]
Rearranging this, we get:
\[ -6\sin^2 \theta - \sin \theta + 3 - 2 = 0 \]
This simplifies to:
\[ -6\sin^2 \theta - \sin \theta + 1 = 0 \]
Multiplying through by -1 to make the leading coefficient positive:
\[ 6\sin^2 \theta + \sin \theta - 1 = 0 \]

### Step 3: Factor the quadratic equation
Now we need to factor the quadratic equation:
\[ 6\sin^2 \theta + 3\sin \theta - 2\sin \theta - 1 = 0 \]
Grouping the terms:
\[ (6\sin^2 \theta + 3\sin \theta) + (-2\sin \theta - 1) = 0 \]
Factoring gives:
\[ 3\sin \theta(2\sin \theta + 1) - 1(2\sin \theta + 1) = 0 \]
This can be factored as:
\[ (3\sin \theta - 1)(2\sin \theta + 1) = 0 \]

### Step 4: Solve for \(\sin \theta\)
Setting each factor to zero gives us two equations:
1. \(3\sin \theta - 1 = 0\)
2. \(2\sin \theta + 1 = 0\)

From the first equation:
\[ 3\sin \theta = 1 \implies \sin \theta = \frac{1}{3} \]

From the second equation:
\[ 2\sin \theta = -1 \implies \sin \theta = -\frac{1}{2} \]

### Step 5: Find the general solutions
For \(\sin \theta = \frac{1}{3}\):
\[ \theta = \sin^{-1}\left(\frac{1}{3}\right) + 2n\pi \quad \text{or} \quad \theta = \pi - \sin^{-1}\left(\frac{1}{3}\right) + 2n\pi \]

For \(\sin \theta = -\frac{1}{2}\):
\[ \theta = \frac{7\pi}{6} + 2n\pi \quad \text{or} \quad \theta = \frac{11\pi}{6} + 2n\pi \]

### Final Solution
Combining both sets of solutions, we have:
\[ \theta = n\pi + (-1)^n \sin^{-1}\left(\frac{1}{3}\right) \quad \text{and} \quad \theta = \frac{7\pi}{6} + 2n\pi \quad \text{or} \quad \theta = \frac{11\pi}{6} + 2n\pi \]