Solve `3tan^(2) theta-2 sin theta=0`.

Explanation

To solve the equation \(3\tan^2 \theta - 2\sin \theta = 0\), we can follow these steps:

### Step 1: Rewrite the equation
Starting with the original equation:
\[ 3\tan^2 \theta - 2\sin \theta = 0 \]
We can rearrange it to:
\[ 3\tan^2 \theta = 2\sin \theta \]

### Step 2: Use the identity for \(\tan^2 \theta\)
We know that:
\[ \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} \]
Substituting this into the equation gives:
\[ 3\frac{\sin^2 \theta}{\cos^2 \theta} = 2\sin \theta \]

### Step 3: Clear the fraction
Multiplying both sides by \(\cos^2 \theta\) (assuming \(\cos \theta \neq 0\)):
\[ 3\sin^2 \theta = 2\sin \theta \cos^2 \theta \]

### Step 4: Use the Pythagorean identity
We know that \(\cos^2 \theta = 1 - \sin^2 \theta\). Substituting this gives:
\[ 3\sin^2 \theta = 2\sin \theta (1 - \sin^2 \theta) \]
Expanding the right side:
\[ 3\sin^2 \theta = 2\sin \theta - 2\sin^3 \theta \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ 2\sin^3 \theta + 3\sin^2 \theta - 2\sin \theta = 0 \]

### Step 6: Factor out \(\sin \theta\)
Factoring out \(\sin \theta\):
\[ \sin \theta (2\sin^2 \theta + 3\sin \theta - 2) = 0 \]

### Step 7: Solve for \(\sin \theta = 0\)
Setting \(\sin \theta = 0\) gives:
\[ \theta = n\pi \quad (n \in \mathbb{Z}) \]

### Step 8: Solve the quadratic equation
Now, we need to solve the quadratic equation:
\[ 2\sin^2 \theta + 3\sin \theta - 2 = 0 \]
Using the quadratic formula:
\[ \sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 2\), \(b = 3\), and \(c = -2\):
\[ \sin \theta = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} \]
Calculating the discriminant:
\[ \sin \theta = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm 5}{4} \]

### Step 9: Find the values of \(\sin \theta\)
Calculating the two possible values:
1. \(\sin \theta = \frac{2}{4} = \frac{1}{2}\)
2. \(\sin \theta = \frac{-8}{4} = -2\) (not possible since \(\sin \theta\) must be between -1 and 1)

### Step 10: Solve for \(\theta\)
For \(\sin \theta = \frac{1}{2}\), we know:
\[ \theta = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad \theta = \frac{5\pi}{6} + 2k\pi \quad (k \in \mathbb{Z}) \]

### Final Solution
Thus, the complete solution set is:
\[ \theta = n\pi \quad (n \in \mathbb{Z}) \quad \text{and} \quad \theta = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad \theta = \frac{5\pi}{6} + 2k\pi \quad (k \in \mathbb{Z}) \]