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: Use the double angle formula for cosine
We know that:
\[ \cos 2\theta = 2\cos^2 \theta - 1 \]
Substituting this into the equation gives us:
\[ 2\cos^2 \theta - 1 = (\sqrt{2} + 1) \left( \cos \theta - \frac{1}{\sqrt{2}} \right) \]

### Step 2: Expand the right-hand side
Expanding the right-hand side:
\[ 2\cos^2 \theta - 1 = (\sqrt{2} + 1) \cos \theta - \frac{\sqrt{2} + 1}{\sqrt{2}} \]
This simplifies to:
\[ 2\cos^2 \theta - 1 = (\sqrt{2} + 1) \cos \theta - (\sqrt{2} + 1) \cdot \frac{1}{\sqrt{2}} \]

### Step 3: Simplify the equation
We can simplify the right-hand side:
\[ \frac{\sqrt{2} + 1}{\sqrt{2}} = 1 + \frac{1}{\sqrt{2}} \]
Thus, we rewrite the equation:
\[ 2\cos^2 \theta - 1 = (\sqrt{2} + 1) \cos \theta - \left(1 + \frac{1}{\sqrt{2}}\right) \]

### Step 4: Rearrange the equation
Rearranging gives us:
\[ 2\cos^2 \theta - (\sqrt{2} + 1) \cos \theta + \left(1 + \frac{1}{\sqrt{2}} - 1\right) = 0 \]
This simplifies to:
\[ 2\cos^2 \theta - (\sqrt{2} + 1) \cos \theta + \frac{1}{\sqrt{2}} = 0 \]

### Step 5: Solve the quadratic equation
Let \( x = \cos \theta \). The equation becomes:
\[ 2x^2 - (\sqrt{2} + 1)x + \frac{1}{\sqrt{2}} = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{(\sqrt{2} + 1) \pm \sqrt{(\sqrt{2} + 1)^2 - 4 \cdot 2 \cdot \frac{1}{\sqrt{2}}}}{2 \cdot 2} \]

### Step 6: Calculate the discriminant
Calculating the discriminant:
\[ (\sqrt{2} + 1)^2 - 4 \cdot 2 \cdot \frac{1}{\sqrt{2}} = 3 + 2\sqrt{2} - \frac{8}{\sqrt{2}} = 3 + 2\sqrt{2} - 4\sqrt{2} = 3 - 2\sqrt{2} \]

### Step 7: Find the roots
Substituting back into the quadratic formula:
\[ x = \frac{(\sqrt{2} + 1) \pm \sqrt{3 - 2\sqrt{2}}}{4} \]

### Step 8: Find the general solutions for \( \theta \)
Now we have two values for \( \cos \theta \):
1. \( \cos \theta = \frac{(\sqrt{2} + 1) + \sqrt{3 - 2\sqrt{2}}}{4} \)
2. \( \cos \theta = \frac{(\sqrt{2} + 1) - \sqrt{3 - 2\sqrt{2}}}{4} \)

For each value of \( \cos \theta \), we can find \( \theta \):
- If \( \cos \theta = k \), then:
\[ \theta = \pm \cos^{-1}(k) + 2n\pi \quad \text{where } n \in \mathbb{Z} \]

### Final Answer
Thus, the general solutions for \( \theta \) will be:
\[ \theta = 2n\pi \pm \cos^{-1}\left(\frac{(\sqrt{2} + 1) + \sqrt{3 - 2\sqrt{2}}}{4}\right) \quad \text{and} \quad \theta = 2n\pi \pm \cos^{-1}\left(\frac{(\sqrt{2} + 1) - \sqrt{3 - 2\sqrt{2}}}{4}\right) \]