If `alpha in [-2pi, 2pi]` and `cos.(alpha)/(2)+sin.(alpha)/(2)=sqrt(2)(cos 36^(@)-sin18^(@))`, then a value of `alpha`

Options

A. `(7pi)/(6)`
B. `(pi)/(6)`
C. `-(5pi)/(6)`
D. `(pi)/(6)`

Explanation

To solve the equation given in the problem, we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \frac{\cos(\alpha)}{2} + \frac{\sin(\alpha)}{2} = \sqrt{2} \left( \cos(36^\circ) - \sin(18^\circ) \right) \]

### Step 2: Factor out \(\frac{1}{2}\)
We can factor out \(\frac{1}{2}\) from the left-hand side:
\[ \frac{1}{2} \left( \cos(\alpha) + \sin(\alpha) \right) = \sqrt{2} \left( \cos(36^\circ) - \sin(18^\circ) \right) \]

### Step 3: Multiply both sides by 2
To eliminate the fraction, multiply both sides by 2:
\[ \cos(\alpha) + \sin(\alpha) = 2\sqrt{2} \left( \cos(36^\circ) - \sin(18^\circ) \right) \]

### Step 4: Calculate the right-hand side
Now we need to calculate \(2\sqrt{2} \left( \cos(36^\circ) - \sin(18^\circ) \right)\):
- Using known values:
- \(\cos(36^\circ) = \frac{\sqrt{5}+1}{4}\)
- \(\sin(18^\circ) = \frac{\sqrt{5}-1}{4}\)

Substituting these values:
\[ 2\sqrt{2} \left( \frac{\sqrt{5}+1}{4} - \frac{\sqrt{5}-1}{4} \right) = 2\sqrt{2} \left( \frac{2}{4} \right) = \sqrt{2} \]

### Step 5: Rewrite the equation
Now we have:
\[ \cos(\alpha) + \sin(\alpha) = \sqrt{2} \]

### Step 6: Use the identity for cosine
We can use the identity:
\[ \cos(\alpha) + \sin(\alpha) = \sqrt{2} \cos\left(\alpha - \frac{\pi}{4}\right) \]
Thus, we can rewrite the equation as:
\[ \sqrt{2} \cos\left(\alpha - \frac{\pi}{4}\right) = \sqrt{2} \]

### Step 7: Divide by \(\sqrt{2}\)
Dividing both sides by \(\sqrt{2}\):
\[ \cos\left(\alpha - \frac{\pi}{4}\right) = 1 \]

### Step 8: Solve for \(\alpha\)
The general solution for \(\cos(x) = 1\) is:
\[ x = 2n\pi \quad \text{for } n \in \mathbb{Z} \]
Thus:
\[ \alpha - \frac{\pi}{4} = 2n\pi \]
This gives:
\[ \alpha = 2n\pi + \frac{\pi}{4} \]

### Step 9: Find specific values of \(\alpha\)
Now we need to find values of \(\alpha\) within the range \([-2\pi, 2\pi]\):
1. For \(n = -1\):
\[ \alpha = -2\pi + \frac{\pi}{4} = -\frac{8\pi}{4} + \frac{\pi}{4} = -\frac{7\pi}{4} \]
2. For \(n = 0\):
\[ \alpha = 0 + \frac{\pi}{4} = \frac{\pi}{4} \]
3. For \(n = 1\):
\[ \alpha = 2\pi + \frac{\pi}{4} = \frac{8\pi}{4} + \frac{\pi}{4} = \frac{9\pi}{4} \quad (\text{not in range}) \]

### Step 10: Final values of \(\alpha\)
The valid solutions for \(\alpha\) in the range \([-2\pi, 2\pi]\) are:
\[ \alpha = -\frac{7\pi}{4}, \quad \alpha = \frac{\pi}{4} \]