Solve the equation for ` 0 le x le 2pi.`
`2 sin ^(2) theta = 3 cos theta `

Answer

To solve the equation \( 2 \sin^2 \theta = 3 \cos \theta \) for \( 0 \leq \theta \leq 2\pi \), we can follow these steps:

### Step 1: Rewrite the equation using the Pythagorean identity
We know that \( \sin^2 \theta = 1 - \cos^2 \theta \). Therefore, we can rewrite the equation as:
\[ 2(1 - \cos^2 \theta) = 3 \cos \theta \]

### Step 2: Expand and rearrange the equation
Expanding the left side gives:
\[ 2 - 2 \cos^2 \theta = 3 \cos \theta \]
Now, rearranging the equation leads to:
\[ 2 \cos^2 \theta + 3 \cos \theta - 2 = 0 \]

### Step 3: Form a quadratic equation
We have a quadratic equation in terms of \( \cos \theta \):
\[ 2 \cos^2 \theta + 3 \cos \theta - 2 = 0 \]

### Step 4: Factor the quadratic equation
To factor the quadratic, we can look for two numbers that multiply to \( 2 \times -2 = -4 \) and add to \( 3 \). The numbers \( 4 \) and \( -1 \) work:
\[ 2 \cos^2 \theta + 4 \cos \theta - \cos \theta - 2 = 0 \]
Grouping the terms gives:
\[ (2 \cos^2 \theta + 4 \cos \theta) + (-\cos \theta - 2) = 0 \]
Factoring by grouping:
\[ 2 \cos \theta (\cos \theta + 2) - 1(\cos \theta + 2) = 0 \]
This can be factored as:
\[ (2 \cos \theta - 1)(\cos \theta + 2) = 0 \]

### Step 5: Solve for \( \cos \theta \)
Setting each factor to zero gives:
1. \( 2 \cos \theta - 1 = 0 \)
2. \( \cos \theta + 2 = 0 \)

For the first equation:
\[ 2 \cos \theta = 1 \implies \cos \theta = \frac{1}{2} \]
For the second equation:
\[ \cos \theta = -2 \]
Since \( \cos \theta \) cannot be less than -1, we discard this solution.

### Step 6: Find the angles for \( \cos \theta = \frac{1}{2} \)
The angles where \( \cos \theta = \frac{1}{2} \) in the range \( 0 \leq \theta \leq 2\pi \) are:
\[ \theta = \frac{\pi}{3}, \quad \text{and} \quad \theta = \frac{5\pi}{3} \]

### Final Solution
Thus, the solutions to the equation \( 2 \sin^2 \theta = 3 \cos \theta \) for \( 0 \leq \theta \leq 2\pi \) are:
\[ \theta = \frac{\pi}{3}, \quad \frac{5\pi}{3} \]