If `2 sin^2 theta = 3cos theta, 0 le theta le 2 pi, theta` =

Options

A. `pi//6, 5pi//6`
B. `pi//3, 2pi//3`
C. `pi//3, 5pi//3`
D. `pi//2, pi`

Answer

To solve the equation \( 2 \sin^2 \theta = 3 \cos \theta \) for \( 0 \leq \theta \leq 2\pi \), we will 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:
\[ 2(1 - \cos^2 \theta) = 3 \cos \theta \]
Expanding this gives:
\[ 2 - 2 \cos^2 \theta = 3 \cos \theta \]

### Step 2: Rearrange the equation
Rearranging the equation leads to:
\[ 2 \cos^2 \theta + 3 \cos \theta - 2 = 0 \]

### Step 3: Solve the quadratic equation
Let \( x = \cos \theta \). The equation becomes:
\[ 2x^2 + 3x - 2 = 0 \]
We can solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2, b = 3, c = -2 \):
\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} \]
Calculating the discriminant:
\[ 3^2 - 4 \cdot 2 \cdot (-2) = 9 + 16 = 25 \]
So, we have:
\[ x = \frac{-3 \pm 5}{4} \]
Calculating the two possible values:
1. \( x = \frac{2}{4} = \frac{1}{2} \)
2. \( x = \frac{-8}{4} = -2 \) (not valid since \( \cos \theta \) must be between -1 and 1)

### Step 4: Find the angle corresponding to \( \cos \theta = \frac{1}{2} \)
The solutions for \( \theta \) when \( \cos \theta = \frac{1}{2} \) are:
\[ \theta = \frac{\pi}{3} \quad \text{and} \quad \theta = \frac{5\pi}{3} \]

### Step 5: List all solutions in the given interval
Since we are looking for solutions in the interval \( 0 \leq \theta \leq 2\pi \), we have:
\[ \theta = \frac{\pi}{3}, \quad \frac{5\pi}{3} \]

Thus, the final solutions are:
\[ \theta = \frac{\pi}{3}, \quad \frac{5\pi}{3} \]