Solutions of the equations `(2cosx-1)(3cosx+4)=0` is `[0,2pi]` is:

Options

A. `pi/3`
B. `pi/3,(5pi)/(3)`
C. `pi/3, (5pi)/(3), cos^(-1)(-4/3)`
D. None of these

Answer

To solve the equation \((2\cos x - 1)(3\cos x + 4) = 0\) in the interval \([0, 2\pi]\), we can follow these steps:

### Step 1: Set each factor to zero
The equation can be split into two separate equations:
1. \(2\cos x - 1 = 0\)
2. \(3\cos x + 4 = 0\)

### 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 angles for \(\cos x = \frac{1}{2}\)
The cosine function equals \(\frac{1}{2}\) at the angles:
\[ x = \frac{\pi}{3} \quad \text{(1st quadrant)} \]
and
\[ x = 2\pi - \frac{\pi}{3} = \frac{6\pi}{3} - \frac{\pi}{3} = \frac{5\pi}{3} \quad \text{(4th quadrant)} \]

### Step 4: Solve the second equation
Now, let's solve the second equation:
\[ 3\cos x + 4 = 0 \]
Subtract 4 from both sides:
\[ 3\cos x = -4 \]
Now, divide both sides by 3:
\[ \cos x = -\frac{4}{3} \]
Since the cosine function has a range of \([-1, 1]\), \(\cos x = -\frac{4}{3}\) is not a valid solution.

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

### Final Answer
Thus, the solutions of the equation \((2\cos x - 1)(3\cos x + 4) = 0\) in the interval \([0, 2\pi]\) are:
\[ x = \frac{\pi}{3}, \frac{5\pi}{3} \]

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