Solve the general vlaue.
`2cos ^(2) theta - 5 cos theta + 2 = 0`

Answer

To solve the equation \(2 \cos^2 \theta - 5 \cos \theta + 2 = 0\), we can follow these steps:

### Step 1: Rewrite the Equation
We start with the quadratic equation in terms of \(\cos \theta\):
\[ 2 \cos^2 \theta - 5 \cos \theta + 2 = 0 \]

### Step 2: Factor the Quadratic
We need to factor the quadratic expression. We can rewrite \(-5 \cos \theta\) as \(-4 \cos \theta - \cos \theta\):
\[ 2 \cos^2 \theta - 4 \cos \theta - \cos \theta + 2 = 0 \]
Now, we can group the terms:
\[ (2 \cos^2 \theta - 4 \cos \theta) + (- \cos \theta + 2) = 0 \]
Factoring out common terms gives:
\[ 2 \cos \theta (\cos \theta - 2) - 1(\cos \theta - 2) = 0 \]
This can be factored as:
\[ (2 \cos \theta - 1)(\cos \theta - 2) = 0 \]

### Step 3: Set Each Factor to Zero
Now, we set each factor equal to zero:
1. \(2 \cos \theta - 1 = 0\)
2. \(\cos \theta - 2 = 0\)

### Step 4: Solve Each Equation
**For the first equation:**
\[ 2 \cos \theta - 1 = 0 \implies \cos \theta = \frac{1}{2} \]
**For the second equation:**
\[ \cos \theta - 2 = 0 \implies \cos \theta = 2 \]
However, \(\cos \theta = 2\) is not valid since the cosine function only takes values in the range \([-1, 1]\).

### Step 5: Find General Solutions for Valid Solutions
Now, we only consider \(\cos \theta = \frac{1}{2}\). The angles that satisfy this equation are:
\[ \theta = \frac{\pi}{3} + 2n\pi \quad \text{and} \quad \theta = -\frac{\pi}{3} + 2n\pi \]
where \(n\) is any integer.

### Final General Solution
Thus, the general solutions for \(\theta\) are:
\[ \theta = \frac{\pi}{3} + 2n\pi \quad \text{and} \quad \theta = \frac{5\pi}{3} + 2n\pi \quad (n \in \mathbb{Z}) \]