Which of the following sets can be the subset of the general solution of `1+c a s3x=2cos2x(n in Z)?` `npi+pi/3` (b) `npi+pi/6` `npi-pi/6` (d) `2npi`

Options

A. `n pi+pi/3`
B. `n pi+pi/6`
C. `n pi- pi/6`
D. `2 n pi`

Explanation

To solve the equation \(1 + \cos 3x = 2 \cos 2x\) and determine which of the given sets can be a subset of the general solution, we can follow these steps:

### Step 1: Rewrite the equation using trigonometric identities
The equation given is:
\[ 1 + \cos 3x = 2 \cos 2x \]
Using the trigonometric identities, we know:
\[ \cos 3x = 4 \cos^3 x - 3 \cos x \]
\[ \cos 2x = 2 \cos^2 x - 1 \]
Substituting these into the equation gives:
\[ 1 + (4 \cos^3 x - 3 \cos x) = 2(2 \cos^2 x - 1) \]

### Step 2: Simplify the equation
Now, simplifying the equation:
\[ 1 + 4 \cos^3 x - 3 \cos x = 4 \cos^2 x - 2 \]
Rearranging terms:
\[ 4 \cos^3 x - 4 \cos^2 x - 3 \cos x + 3 = 0 \]

### Step 3: Factor the equation
We can factor this equation. Notice that we can group terms:
\[ 4 \cos^3 x - 4 \cos^2 x - 3 \cos x + 3 = 0 \]
Factoring out common terms:
\[ (4 \cos^2 x - 3)(\cos x - 1) = 0 \]

### Step 4: Solve the factored equations
Now we have two equations to solve:
1. \(4 \cos^2 x - 3 = 0\)
2. \(\cos x - 1 = 0\)

#### For the first equation:
\[ 4 \cos^2 x = 3 \implies \cos^2 x = \frac{3}{4} \implies \cos x = \pm \frac{\sqrt{3}}{2} \]
This gives us:
\[ x = n\pi \pm \frac{\pi}{6} \quad (n \in \mathbb{Z}) \]

#### For the second equation:
\[ \cos x = 1 \implies x = 2n\pi \quad (n \in \mathbb{Z}) \]

### Step 5: Write the general solution
Thus, the general solutions we have are:
1. \(x = n\pi + \frac{\pi}{6}\)
2. \(x = n\pi - \frac{\pi}{6}\)
3. \(x = 2n\pi\)

### Step 6: Analyze the options
Now we can analyze the given options:
- (a) \(n\pi + \frac{\pi}{3}\) - **Not a solution**
- (b) \(n\pi + \frac{\pi}{6}\) - **Is a solution**
- (c) \(n\pi - \frac{\pi}{6}\) - **Is a solution**
- (d) \(2n\pi\) - **Is a solution**

### Conclusion
The sets that can be subsets of the general solution are:
- (b) \(n\pi + \frac{\pi}{6}\)
- (c) \(n\pi - \frac{\pi}{6}\)
- (d) \(2n\pi\)