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The equation given is:

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

We need to solve for all solutions \(x\) in the interval \([0^\circ, 360^\circ]\).

### Step-by-Step Solution:

1. **Rewrite the equation:**
   Move all terms to one side to make the equation easier to factor:
   
   \[
   2 \sin^2(x) + \sin(x) - 1 = 0
   \]

2. **Factor the quadratic equation:**
   This is a quadratic equation in terms of \(\sin(x)\). We can factor it:

   \[
   (2 \sin(x) - 1)(\sin(x) + 1) = 0
   \]

3. **Solve each factor:**

   a. **First factor:**
      \[
      2 \sin(x) - 1 = 0
      \]
      Solving for \(\sin(x)\):
      \[
      \sin(x) = \frac{1}{2}
      \]
      The general solutions for \(\sin(x) = \frac{1}{2}\) in \([0^\circ, 360^\circ]\) are:
      \[
      x = 30^\circ, 150^\circ
      \]
   
   b. **Second factor:**
      \[
      \sin(x) + 1 = 0
      \]
      Solving for \(\sin(x)\):
      \[
      \sin(x) = -1
      \]
      The general solution for \(\sin(x) = -1\) in \([0^\circ, 360^\circ]\) is:
      \[
      x = 270^\circ
      \]

### Final Solutions:
The solutions for \(x\) in \([0^\circ, 360^\circ]\) are:
\[
x = 30^\circ, 150^\circ, 270^\circ
\]

Would you like a detailed explanation of any step? 

### Related questions:
1. What is the significance of the range \([0^\circ, 360^\circ]\) in trigonometric equations?
2. How do you factor quadratic trigonometric equations?
3. What are the general solutions for \(\sin(x) = \frac{1}{2}\) and \(\sin(x) = -1\)?
4. How can you apply the unit circle to find sine values at specific angles?
5. What method is used to solve trigonometric equations that can't be factored easily?

**Tip:** Always check the interval in which you need to find the solution to avoid missing any valid solutions.

Solve, finding all solutions in [0, 2π] or [0°, 360°]: 2sin²(x) + sin(x) = 1

Learn how to solve the trigonometric equation 2sin²(x) + sin(x) = 1 by factoring a quadratic equation and finding solutions in the interval [0°, 360°]. This problem covers key trigonometric concepts like the sine function and unit circle.

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