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The problem is to solve the trigonometric equation:

\[
2 \sin^2(x) + \sin(x) = 1
\]

### Step 1: Set Up the Equation
Let’s start by letting \( y = \sin(x) \). Substituting \( y \) into the equation gives:

\[
2y^2 + y - 1 = 0
\]

### Step 2: Solve the Quadratic Equation
This is a quadratic equation in terms of \( y \). We can solve it using the quadratic formula:

\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \( a = 2 \), \( b = 1 \), and \( c = -1 \). Plugging in these values:

\[
y = \frac{-1 \pm \sqrt{1^2 - 4(2)(-1)}}{2(2)}
\]
\[
y = \frac{-1 \pm \sqrt{1 + 8}}{4}
\]
\[
y = \frac{-1 \pm \sqrt{9}}{4}
\]
\[
y = \frac{-1 \pm 3}{4}
\]

This gives two solutions:

1. \( y = \frac{2}{4} = \frac{1}{2} \)
2. \( y = \frac{-4}{4} = -1 \)

### Step 3: Find \( x \) for Each Solution
Now, solve for \( x \) in each case.

1. **For \( \sin(x) = \frac{1}{2} \):**
   - \( x = \frac{\pi}{6} \) or \( x = \frac{5\pi}{6} \) in the interval \([0, 2\pi]\).

2. **For \( \sin(x) = -1 \):**
   - \( x = \frac{3\pi}{2} \) in the interval \([0, 2\pi]\).

### Step 4: List All Solutions
The solutions in \([0, 2\pi]\) are:

\[
x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}
\]

These are the exact solutions in radians.

### Step 5: Convert Solutions to Degrees
If you prefer degrees, the solutions are:

\[
x = 30^\circ, 150^\circ, 270^\circ
\]

### Final Answer
- In degrees: \( x = 30^\circ, 150^\circ, 270^\circ \)
- In radians: \( x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2} \)

Would you like more details or have any questions about this solution?

### Relative Questions:
1. How do you convert between radians and degrees?
2. What are the properties of the sine function on the unit circle?
3. How can you verify if solutions to trigonometric equations are correct?
4. Why is the quadratic formula useful in solving trigonometric equations?
5. What is the general solution for \( \sin(x) = k \)?

**Tip:** When solving trigonometric equations, always check the interval given to make sure all solutions fit within that range.

Solve 2sin^2(x) + sin(x) = 1, finding all solutions in [0, 2π] or [0°, 360°].

Learn how to solve the trigonometric equation 2sin^2(x) + sin(x) = 1 by converting it into a quadratic form. This step-by-step solution covers concepts like the quadratic formula and the properties of the sine function. Suitable for Grades 9-12.

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