If `3tan^(2)theta-2sintheta=0`, then general value of `theta` is:

Options

A. `npi`
B. `(npi)/(2)`
C. `(npi)/(3)`
D. `(npi)/(4)`

Answer

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

### Step 1: Rewrite the equation
Starting with the given equation:
\[ 3 \tan^2 \theta - 2 \sin \theta = 0 \]
We can rearrange it to isolate \(\tan^2 \theta\):
\[ 3 \tan^2 \theta = 2 \sin \theta \]
Thus,
\[ \tan^2 \theta = \frac{2}{3} \sin \theta \]

### Step 2: Use the identity for \(\tan^2 \theta\)
Recall the identity:
\[ \tan^2 \theta = \sec^2 \theta - 1 \]
Substituting this into our equation gives:
\[ \sec^2 \theta - 1 = \frac{2}{3} \sin \theta \]
This simplifies to:
\[ \sec^2 \theta = \frac{2}{3} \sin \theta + 1 \]

### Step 3: Express \(\sec^2 \theta\) in terms of \(\sin \theta\)
Since \(\sec^2 \theta = \frac{1}{\cos^2 \theta}\), we can write:
\[ \frac{1}{\cos^2 \theta} = \frac{2}{3} \sin \theta + 1 \]
Multiplying both sides by \(\cos^2 \theta\) gives:
\[ 1 = \left(\frac{2}{3} \sin \theta + 1\right) \cos^2 \theta \]

### Step 4: Rearranging the equation
Rearranging the equation leads to:
\[ \left(\frac{2}{3} \sin \theta + 1\right) \cos^2 \theta = 1 \]
This can be rewritten as:
\[ \cos^2 \theta = \frac{1}{\frac{2}{3} \sin \theta + 1} \]

### Step 5: Use the Pythagorean identity
Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\), we can express \(\cos^2 \theta\) in terms of \(\sin \theta\):
\[ \cos^2 \theta = 1 - \sin^2 \theta \]
Substituting this into the previous equation gives:
\[ 1 - \sin^2 \theta = \frac{1}{\frac{2}{3} \sin \theta + 1} \]

### Step 6: Cross-multiply and simplify
Cross-multiplying yields:
\[ (1 - \sin^2 \theta)(\frac{2}{3} \sin \theta + 1) = 1 \]
Expanding this gives:
\[ \frac{2}{3} \sin \theta - \frac{2}{3} \sin^3 \theta + 1 - \sin^2 \theta = 1 \]
Simplifying leads to:
\[ \frac{2}{3} \sin \theta - \frac{2}{3} \sin^3 \theta - \sin^2 \theta = 0 \]
Multiplying through by 3 to eliminate the fraction:
\[ 2 \sin \theta - 2 \sin^3 \theta - 3 \sin^2 \theta = 0 \]

### Step 7: Factor out \(\sin \theta\)
Factoring out \(\sin \theta\):
\[ \sin \theta (2 - 2 \sin^2 \theta - 3 \sin \theta) = 0 \]
This gives us two cases:
1. \(\sin \theta = 0\)
2. \(2 - 2 \sin^2 \theta - 3 \sin \theta = 0\)

### Step 8: Solve for \(\sin \theta = 0\)
From \(\sin \theta = 0\):
\[ \theta = n\pi \quad (n \in \mathbb{Z}) \]

### Step 9: Solve the quadratic equation
For the quadratic equation \(2 - 2 \sin^2 \theta - 3 \sin \theta = 0\), we can rearrange it to:
\[ 2 \sin^2 \theta + 3 \sin \theta - 2 = 0 \]
Using the quadratic formula:
\[ \sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} \]
Calculating the discriminant:
\[ = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm 5}{4} \]
This gives:
\[ \sin \theta = \frac{2}{4} = \frac{1}{2} \quad \text{or} \quad \sin \theta = \frac{-8}{4} = -2 \quad (\text{not possible}) \]

### Step 10: Find \(\theta\) for \(\sin \theta = \frac{1}{2}\)
From \(\sin \theta = \frac{1}{2}\):
\[ \theta = \frac{\pi}{6} + 2n\pi \quad \text{or} \quad \theta = \frac{5\pi}{6} + 2n\pi \]

### Final General Solution
Combining all solutions, the general values of \(\theta\) are:
\[ \theta = n\pi \quad (n \in \mathbb{Z}) \quad \text{and} \quad \theta = \frac{\pi}{6} + 2n\pi \quad \text{or} \quad \theta = \frac{5\pi}{6} + 2n\pi \]