To solve the given problem, we need to analyze the two statements and verify their truthfulness step by step.
### Step 1: Analyze Statement 2
**Statement 2**: If \( 2\cos^2\theta - 3\sin\theta = 0 \), then \( \theta \) does not lie in the third or fourth quadrant.
1. Start with the equation:
\[
2\cos^2\theta - 3\sin\theta = 0
\]
Rearranging gives:
\[
2\cos^2\theta = 3\sin\theta
\]
2. Recall that in the third and fourth quadrants, \( \sin\theta \) is negative. Therefore, if \( \sin\theta < 0 \), then \( 3\sin\theta < 0 \).
3. Since \( \cos^2\theta \) is always non-negative (as it is a square), \( 2\cos^2\theta \) is also non-negative. Thus, we have:
\[
2\cos^2\theta \geq 0 \quad \text{and} \quad 3\sin\theta < 0
\]
This implies:
\[
2\cos^2\theta - 3\sin\theta > 0
\]
Therefore, the equation \( 2\cos^2\theta - 3\sin\theta = 0 \) cannot hold true in the third or fourth quadrant.
**Conclusion for Statement 2**: True. \( \theta \) does not lie in the third or fourth quadrant.
### Step 2: Analyze Statement 1
**Statement 1**: The number of solutions of the simultaneous system of equations
\[
2\sin^2\theta - \cos 2\theta = 0
\]
\[
2\cos^2\theta - 3\sin\theta = 0
\]
in the interval \([0, 2\pi]\) is two.
1. Start with the first equation:
\[
2\sin^2\theta - \cos 2\theta = 0
\]
Using the identity \( \cos 2\theta = \cos^2\theta - \sin^2\theta \), we can rewrite this as:
\[
2\sin^2\theta - (\cos^2\theta - \sin^2\theta) = 0
\]
Simplifying gives:
\[
2\sin^2\theta + \sin^2\theta - \cos^2\theta = 0
\]
or
\[
3\sin^2\theta - \cos^2\theta = 0
\]
2. Substitute \( \cos^2\theta = 1 - \sin^2\theta \):
\[
3\sin^2\theta - (1 - \sin^2\theta) = 0
\]
which simplifies to:
\[
4\sin^2\theta - 1 = 0
\]
Thus:
\[
\sin^2\theta = \frac{1}{4}
\]
Therefore:
\[
\sin\theta = \frac{1}{2} \quad \text{or} \quad \sin\theta = -\frac{1}{2}
\]
3. The solutions for \( \sin\theta = \frac{1}{2} \) in the interval \([0, 2\pi]\) are:
\[
\theta = \frac{\pi}{6}, \quad \frac{5\pi}{6}
\]
4. The solutions for \( \sin\theta = -\frac{1}{2} \) in the interval \([0, 2\pi]\) are:
\[
\theta = \frac{7\pi}{6}, \quad \frac{11\pi}{6}
\]
5. Now, we have four solutions:
\[
\frac{\pi}{6}, \quad \frac{5\pi}{6}, \quad \frac{7\pi}{6}, \quad \frac{11\pi}{6}
\]
6. Next, check the second equation:
\[
2\cos^2\theta - 3\sin\theta = 0
\]
Rearranging gives:
\[
2\cos^2\theta = 3\sin\theta
\]
Substitute \( \cos^2\theta = 1 - \sin^2\theta \):
\[
2(1 - \sin^2\theta) = 3\sin\theta
\]
This leads to:
\[
2 - 2\sin^2\theta - 3\sin\theta = 0
\]
Rearranging gives:
\[
2\sin^2\theta + 3\sin\theta - 2 = 0
\]
7. Solving this quadratic equation using the quadratic formula:
\[
\sin\theta = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2}
\]
Simplifying gives:
\[
\sin\theta = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm 5}{4}
\]
Thus:
\[
\sin\theta = \frac{2}{4} = \frac{1}{2} \quad \text{or} \quad \sin\theta = \frac{-8}{4} = -2 \quad (\text{not valid})
\]
8. The valid solution from this equation is \( \sin\theta = \frac{1}{2} \), which we already found gives us two angles.
### Conclusion for Statement 1
The total number of solutions in the interval \([0, 2\pi]\) is four, not two. Therefore, **Statement 1 is false**.
### Final Conclusion
- **Statement 1**: False
- **Statement 2**: True
