The number of solutions of the pair of equations `2sin^2theta-cos2theta=0` `2cos^2theta-3sintheta=0` in the interval `[0,2pi]` is

Options

A. 0
B. 1
C. 2
D. 4

Explanation

To find the number of solutions of the given pair of equations in the interval \([0, 2\pi]\), we will solve each equation step by step.

### Step 1: Solve the first equation
The first equation is:
\[ 2\sin^2\theta - \cos 2\theta = 0 \]
Using the identity \(\cos 2\theta = 1 - 2\sin^2\theta\), we can substitute:
\[ 2\sin^2\theta - (1 - 2\sin^2\theta) = 0 \]
This simplifies to:
\[ 2\sin^2\theta - 1 + 2\sin^2\theta = 0 \]
Combining like terms gives:
\[ 4\sin^2\theta - 1 = 0 \]
Rearranging, we find:
\[ 4\sin^2\theta = 1 \]
Dividing both sides by 4:
\[ \sin^2\theta = \frac{1}{4} \]
Taking the square root:
\[ \sin\theta = \pm \frac{1}{2} \]

### Step 2: Find angles for \(\sin\theta = \frac{1}{2}\) and \(\sin\theta = -\frac{1}{2}\)
1. For \(\sin\theta = \frac{1}{2}\):
- The solutions in the interval \([0, 2\pi]\) are:
\[ \theta = \frac{\pi}{6}, \quad \frac{5\pi}{6} \]

2. For \(\sin\theta = -\frac{1}{2}\):
- The solutions in the interval \([0, 2\pi]\) are:
\[ \theta = \frac{7\pi}{6}, \quad \frac{11\pi}{6} \]

Thus, from the first equation, we have four solutions:
\[ \theta = \frac{\pi}{6}, \quad \frac{5\pi}{6}, \quad \frac{7\pi}{6}, \quad \frac{11\pi}{6} \]

### Step 3: Solve the second equation
The second equation is:
\[ 2\cos^2\theta - 3\sin\theta = 0 \]
Using the identity \(\cos^2\theta = 1 - \sin^2\theta\), we can rewrite the equation:
\[ 2(1 - \sin^2\theta) - 3\sin\theta = 0 \]
Expanding gives:
\[ 2 - 2\sin^2\theta - 3\sin\theta = 0 \]
Rearranging leads to:
\[ 2\sin^2\theta + 3\sin\theta - 2 = 0 \]

### Step 4: Factor the quadratic equation
To factor \(2\sin^2\theta + 3\sin\theta - 2 = 0\), we look for two numbers that multiply to \(-4\) (the product of \(2\) and \(-2\)) and add to \(3\). The numbers \(4\) and \(-1\) work:
\[ (2\sin\theta - 1)(\sin\theta + 2) = 0 \]

### Step 5: Solve each factor
1. From \(2\sin\theta - 1 = 0\):
\[ \sin\theta = \frac{1}{2} \]
The solutions are:
\[ \theta = \frac{\pi}{6}, \quad \frac{5\pi}{6} \]

2. From \(\sin\theta + 2 = 0\):
\[ \sin\theta = -2 \]
This is not possible since \(\sin\theta\) must be between \(-1\) and \(1\).

Thus, from the second equation, we have two solutions:
\[ \theta = \frac{\pi}{6}, \quad \frac{5\pi}{6} \]

### Step 6: Find common solutions
Now, we find the common solutions from both equations:
- From the first equation: \(\frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}\)
- From the second equation: \(\frac{\pi}{6}, \frac{5\pi}{6}\)

The common solutions are:
\[ \frac{\pi}{6}, \quad \frac{5\pi}{6} \]

### Conclusion
The number of solutions of the pair of equations in the interval \([0, 2\pi]\) is:
\[ \boxed{2} \]