If `6sin^(2)theta - sin theta = 1` and `0 le theta le pi`, what is the value of sin `theta` ?

Options

A. `(1)/(6)`
B. `(1)/(3)`
C. `(1)/(2)`
D. 19

Explanation

To solve the equation \( 6\sin^2\theta - \sin\theta = 1 \) for \( \theta \) in the interval \( 0 \leq \theta \leq \pi \), we can follow these steps:

### Step 1: Substitute \( \sin \theta \) with \( x \)
Let \( x = \sin \theta \). Then, we can rewrite the equation as:
\[ 6x^2 - x = 1 \]

### Step 2: Rearrange the equation
Rearranging the equation gives us:
\[ 6x^2 - x - 1 = 0 \]

### Step 3: Factor the quadratic equation
Next, we need to factor the quadratic equation. We can do this by finding two numbers that multiply to \( 6 \times (-1) = -6 \) and add to \( -1 \). The numbers \( -3 \) and \( 2 \) work because:
\[ -3 + 2 = -1 \quad \text{and} \quad -3 \times 2 = -6 \]
Thus, we can rewrite the equation as:
\[ 6x^2 - 3x + 2x - 1 = 0 \]

### Step 4: Group the terms
Now, we can group the terms:
\[ (6x^2 - 3x) + (2x - 1) = 0 \]

### Step 5: Factor by grouping
Factoring out the common terms gives us:
\[ 3x(2x - 1) + 1(2x - 1) = 0 \]
This can be factored further as:
\[ (2x - 1)(3x + 1) = 0 \]

### Step 6: Solve for \( x \)
Setting each factor to zero gives us:
1. \( 2x - 1 = 0 \) → \( x = \frac{1}{2} \)
2. \( 3x + 1 = 0 \) → \( x = -\frac{1}{3} \)

### Step 7: Determine valid solutions
Since \( x = \sin \theta \) and \( \sin \theta \) must be in the range \([-1, 1]\), we check the values:
- \( x = \frac{1}{2} \) is valid.
- \( x = -\frac{1}{3} \) is not valid since sine is positive in the interval \( 0 \leq \theta \leq \pi \).

### Conclusion
Thus, the only valid solution for \( \sin \theta \) is:
\[ \sin \theta = \frac{1}{2} \]