If `3 sin 2 theta = 2 sin 3 theta` and `0 lt theta lt pi`, then sin `theta = `

Options

A. `sqrt(2)//3`
B. `sqrt(3)//sqrt(5)`
C. `sqrt(15)//4`
D. `sqrt(2)//sqrt(5)`

Answer

To solve the equation \(3 \sin 2\theta = 2 \sin 3\theta\) for \(0 < \theta < \pi\), we can follow these steps:

### Step 1: Use the Double Angle and Triple Angle Formulas
We know that:
\[ \sin 2\theta = 2 \sin \theta \cos \theta \]
\[ \sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta \]

Substituting these into the original equation gives:
\[ 3(2 \sin \theta \cos \theta) = 2(3 \sin \theta - 4 \sin^3 \theta) \]
This simplifies to:
\[ 6 \sin \theta \cos \theta = 6 \sin \theta - 8 \sin^3 \theta \]

### Step 2: Rearranging the Equation
Rearranging the equation, we get:
\[ 6 \sin \theta \cos \theta - 6 \sin \theta + 8 \sin^3 \theta = 0 \]

### Step 3: Factor Out \(\sin \theta\)
Factoring out \(\sin \theta\) gives:
\[ \sin \theta (6 \cos \theta - 6 + 8 \sin^2 \theta) = 0 \]

This leads to two cases:
1. \(\sin \theta = 0\)
2. \(6 \cos \theta - 6 + 8 \sin^2 \theta = 0\)

### Step 4: Solve for \(\sin \theta = 0\)
The first case \(\sin \theta = 0\) gives:
\[ \theta = 0 \quad \text{(not in the interval \(0 < \theta < \pi\))} \]

### Step 5: Solve the Second Equation
For the second case, we can substitute \(\sin^2 \theta = 1 - \cos^2 \theta\):
\[ 6 \cos \theta - 6 + 8(1 - \cos^2 \theta) = 0 \]
This simplifies to:
\[ 6 \cos \theta - 6 + 8 - 8 \cos^2 \theta = 0 \]
\[ -8 \cos^2 \theta + 6 \cos \theta + 2 = 0 \]

### Step 6: Rearranging the Quadratic Equation
Rearranging gives:
\[ 8 \cos^2 \theta - 6 \cos \theta - 2 = 0 \]

### Step 7: Using the Quadratic Formula
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
Here, \(a = 8\), \(b = -6\), and \(c = -2\):
\[ \cos \theta = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 8 \cdot (-2)}}{2 \cdot 8} \]
\[ = \frac{6 \pm \sqrt{36 + 64}}{16} \]
\[ = \frac{6 \pm \sqrt{100}}{16} \]
\[ = \frac{6 \pm 10}{16} \]

### Step 8: Finding the Values of \(\cos \theta\)
Calculating the two possible values:
1. \(\cos \theta = \frac{16}{16} = 1\) (which gives \(\theta = 0\), not valid in our interval)
2. \(\cos \theta = \frac{-4}{16} = -\frac{1}{4}\)

### Step 9: Finding \(\sin \theta\)
Using \(\cos \theta = -\frac{1}{4}\):
\[ \sin^2 \theta = 1 - \left(-\frac{1}{4}\right)^2 = 1 - \frac{1}{16} = \frac{15}{16} \]
Thus,
\[ \sin \theta = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4} \]

### Final Answer
\[ \sin \theta = \frac{\sqrt{15}}{4} \]