If `cos 2theta=(sqrt(2)+1)(cos theta-(1)/(sqrt(2)))`, then the general value of `theta(n in Z)`

Options

A. `2n pi pm (pi)/(6)`
B. `n pi+(pi)/(12)`
C. `n pi+(7pi)/(36)`
D. none of these

Explanation

To solve the equation \( \cos 2\theta = (\sqrt{2}+1)\left(\cos \theta - \frac{1}{\sqrt{2}}\right) \), we will follow these steps:

### Step 1: Substitute for \( \cos 2\theta \)
Using the double angle formula for cosine, we know that:
\[ \cos 2\theta = 2\cos^2 \theta - 1 \]
Substituting this into the equation gives:
\[ 2\cos^2 \theta - 1 = (\sqrt{2}+1)\left(\cos \theta - \frac{1}{\sqrt{2}}\right) \]

### Step 2: Expand the right-hand side
Now, we expand the right-hand side:
\[ (\sqrt{2}+1)\left(\cos \theta - \frac{1}{\sqrt{2}}\right) = (\sqrt{2}+1)\cos \theta - \frac{\sqrt{2}+1}{\sqrt{2}} \]
This simplifies to:
\[ (\sqrt{2}+1)\cos \theta - (1 + \frac{1}{\sqrt{2}}) \]

### Step 3: Rearranging the equation
Now, we rearrange the equation:
\[ 2\cos^2 \theta - 1 = (\sqrt{2}+1)\cos \theta - (1 + \frac{1}{\sqrt{2}}) \]
Bringing everything to one side gives:
\[ 2\cos^2 \theta - (\sqrt{2}+1)\cos \theta + \left(1 + \frac{1}{\sqrt{2}} - 1\right) = 0 \]
This simplifies to:
\[ 2\cos^2 \theta - (\sqrt{2}+1)\cos \theta + \frac{1}{\sqrt{2}} = 0 \]

### Step 4: Solve the quadratic equation
This is a quadratic equation in terms of \( \cos \theta \):
\[ 2x^2 - (\sqrt{2}+1)x + \frac{1}{\sqrt{2}} = 0 \]
where \( x = \cos \theta \).

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- \( a = 2 \)
- \( b = -(\sqrt{2}+1) \)
- \( c = \frac{1}{\sqrt{2}} \)

Calculating the discriminant:
\[ b^2 - 4ac = (\sqrt{2}+1)^2 - 4 \cdot 2 \cdot \frac{1}{\sqrt{2}} \]
Calculating \( (\sqrt{2}+1)^2 = 2 + 2\sqrt{2} + 1 = 3 + 2\sqrt{2} \) and \( 4 \cdot 2 \cdot \frac{1}{\sqrt{2}} = 4\sqrt{2} \):
\[ 3 + 2\sqrt{2} - 4\sqrt{2} = 3 - 2\sqrt{2} \]

### Step 5: Finding the roots
Now substituting back into the quadratic formula:
\[ x = \frac{\sqrt{2}+1 \pm \sqrt{3 - 2\sqrt{2}}}{4} \]
This gives us two values for \( \cos \theta \).

### Step 6: General solutions for \( \theta \)
For each value of \( \cos \theta \), we find the general solutions:
1. If \( \cos \theta = \frac{1}{\sqrt{2}} \), then:
\[ \theta = 2n\pi \pm \frac{\pi}{4}, \quad n \in \mathbb{Z} \]
2. If \( \cos \theta = \frac{1}{2} \), then:
\[ \theta = 2n\pi \pm \frac{\pi}{3}, \quad n \in \mathbb{Z} \]

### Final Answer
The general values of \( \theta \) are:
\[ \theta = 2n\pi \pm \frac{\pi}{4} \quad \text{and} \quad \theta = 2n\pi \pm \frac{\pi}{3}, \quad n \in \mathbb{Z} \]