`{x in R: "cos" 2x + 2"cos"^(2) x = 2}` is equal to

Options

A. `{2npi + (pi)/(3) : n in Z}`
B. `{n pi +-(pi)/(6) : n in Z}`
C. `{n pi + (pi)/(3): n in Z}`
D. `{2n pi-(pi)/(3): n in Z}`

Answer

To solve the equation \( \cos(2x) + 2\cos^2(x) = 2 \), we will follow these steps:

### Step 1: Rewrite the equation
Start with the given equation:
\[ \cos(2x) + 2\cos^2(x) = 2 \]

### Step 2: Use the double angle formula
Recall the double angle formula for cosine:
\[ \cos(2x) = 2\cos^2(x) - 1 \]
Substituting this into the equation gives:
\[ (2\cos^2(x) - 1) + 2\cos^2(x) = 2 \]

### Step 3: Combine like terms
Combine the terms involving \(\cos^2(x)\):
\[ 2\cos^2(x) - 1 + 2\cos^2(x) = 2 \]
This simplifies to:
\[ 4\cos^2(x) - 1 = 2 \]

### Step 4: Isolate the cosine term
Add 1 to both sides:
\[ 4\cos^2(x) = 3 \]

### Step 5: Solve for \(\cos^2(x)\)
Divide both sides by 4:
\[ \cos^2(x) = \frac{3}{4} \]

### Step 6: Take the square root
Taking the square root gives:
\[ \cos(x) = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2} \]

### Step 7: Find the general solutions for \(x\)
The cosine function equals \(\frac{\sqrt{3}}{2}\) at:
\[ x = \frac{\pi}{6} + 2k\pi \quad \text{and} \quad x = \frac{11\pi}{6} + 2k\pi \quad (k \in \mathbb{Z}) \]
The cosine function equals \(-\frac{\sqrt{3}}{2}\) at:
\[ x = \frac{5\pi}{6} + 2k\pi \quad \text{and} \quad x = \frac{7\pi}{6} + 2k\pi \quad (k \in \mathbb{Z}) \]

### Step 8: Combine all solutions
Thus, the complete solution set is:
\[ x = \frac{\pi}{6} + 2k\pi, \quad x = \frac{11\pi}{6} + 2k\pi, \quad x = \frac{5\pi}{6} + 2k\pi, \quad x = \frac{7\pi}{6} + 2k\pi \quad (k \in \mathbb{Z}) \]

### Final Answer
The solution set is:
\[ \{ x \in \mathbb{R} : x = \frac{\pi}{6} + 2k\pi, \frac{11\pi}{6} + 2k\pi, \frac{5\pi}{6} + 2k\pi, \frac{7\pi}{6} + 2k\pi, k \in \mathbb{Z} \} \]