The solution of (2 cos x – 1) (3 + 2 cos x) = 0 in the interval `0 lt= x lt= 2pi` is

Options

A. `pi/3`
B. `pi/3, (5pi)/(3)`
C. `pi/3, (5pi)/(3), cos^(-1) (-3/2) `
D. none of these

Answer

To solve the equation \((2 \cos x - 1)(3 + 2 \cos x) = 0\) in the interval \(0 \leq x \leq 2\pi\), we can break it down into two separate equations:

1. \(2 \cos x - 1 = 0\)
2. \(3 + 2 \cos x = 0\)

### Step 1: Solve the first equation

Starting with the first equation:

\[ 2 \cos x - 1 = 0 \]

Add 1 to both sides:

\[ 2 \cos x = 1 \]

Now, divide both sides by 2:

\[ \cos x = \frac{1}{2} \]

### Step 2: Find the values of \(x\)

The cosine function equals \(\frac{1}{2}\) at specific angles in the unit circle. In the interval \(0 \leq x \leq 2\pi\), the angles where \(\cos x = \frac{1}{2}\) are:

\[ x = \frac{\pi}{3} \quad \text{and} \quad x = \frac{5\pi}{3} \]

### Step 3: Solve the second equation

Now, let's solve the second equation:

\[ 3 + 2 \cos x = 0 \]

Subtract 3 from both sides:

\[ 2 \cos x = -3 \]

Now, divide both sides by 2:

\[ \cos x = -\frac{3}{2} \]

### Step 4: Analyze the second equation

The value \(-\frac{3}{2}\) is not possible for the cosine function, as the range of \(\cos x\) is \([-1, 1]\). Therefore, there are no solutions from this equation.

### Final Solution

The only solutions in the interval \(0 \leq x \leq 2\pi\) are from the first equation:

\[ x = \frac{\pi}{3} \quad \text{and} \quad x = \frac{5\pi}{3} \]

### Summary of Solutions

Thus, the solutions to the equation \((2 \cos x - 1)(3 + 2 \cos x) = 0\) in the interval \(0 \leq x \leq 2\pi\) are:

\[ x = \frac{\pi}{3}, \quad x = \frac{5\pi}{3} \]