The solution set of `(2"cos"x-1)(3+2"cos"x) = 0` in the interval `0 le x le 2 pi`, 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 follow these steps:

### Step 1: Set each factor to zero
We start by setting each factor of the equation to zero:
1. \(2 \cos x - 1 = 0\)
2. \(3 + 2 \cos x = 0\)

### Step 2: Solve the first equation
From 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 3: Find the values of \(x\) for \(\cos x = \frac{1}{2}\)
The cosine function equals \(\frac{1}{2}\) at the angles:
\[ x = \frac{\pi}{3} \quad \text{(1st quadrant)} \]
\[ x = \frac{5\pi}{3} \quad \text{(4th quadrant)} \]

### Step 4: Solve the second equation
Now, we solve the second equation:
\[ 3 + 2 \cos x = 0 \]
Subtract 3 from both sides:
\[ 2 \cos x = -3 \]
Now, divide both sides by 2:
\[ \cos x = -\frac{3}{2} \]
Since the cosine function only takes values in the range \([-1, 1]\), \(\cos x = -\frac{3}{2}\) is not a valid solution.

### Step 5: Compile the solution set
The valid solutions from the first equation are:
\[ x = \frac{\pi}{3}, \quad x = \frac{5\pi}{3} \]

### Final Answer
Thus, the solution set of the equation \((2 \cos x - 1)(3 + 2 \cos x) = 0\) in the interval \(0 \leq x \leq 2\pi\) is:
\[ \left\{ \frac{\pi}{3}, \frac{5\pi}{3} \right\} \]

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