select all solutions to the equation 2sin^2x + sinx-1=0 on the interval [0, 2pi)

To solve the equation \( 2\sin^2x + \sin x - 1 = 0 \) on the interval \( [0, 2\pi) \), we follow these steps:

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### 1. Substitution
Let \( y = \sin x \). The equation becomes:
\[
2y^2 + y - 1 = 0
\]

---

### 2. Solve the quadratic equation
Using the quadratic formula:
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 2 \), \( b = 1 \), and \( c = -1 \), we get:
\[
y = \frac{-1 \pm \sqrt{1^2 - 4(2)(-1)}}{2(2)}
\]
\[
y = \frac{-1 \pm \sqrt{1 + 8}}{4}
\]
\[
y = \frac{-1 \pm \sqrt{9}}{4}
\]
\[
y = \frac{-1 \pm 3}{4}
\]

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### 3. Simplify solutions for \( y \):
\[
y = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2}
\]
\[
y = \frac{-1 - 3}{4} = \frac{-4}{4} = -1
\]

So, \( y = \frac{1}{2} \) or \( y = -1 \).

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### 4. Solve for \( \sin x \):
1. \( \sin x = \frac{1}{2} \):
   \[
   x = \arcsin\left(\frac{1}{2}\right) \implies x = \frac{\pi}{6}, \frac{5\pi}{6}
   \]
   (in quadrants I and II where sine is positive).

2. \( \sin x = -1 \):
   \[
   x = \arcsin(-1) \implies x = \frac{3\pi}{2}
   \]
   (in quadrant III where sine is -1).

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### 5. Final solutions
The solutions to the equation \( 2\sin^2x + \sin x - 1 = 0 \) on the interval \( [0, 2\pi) \) are:
\[
x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}
\]

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Would you like a detailed explanation of these steps? 

Here are 5 related questions for further practice:
1. Solve \( 3\sin^2x - 2\sin x - 1 = 0 \) on \( [0, 2\pi) \).
2. Solve \( \cos^2x - 3\cos x + 2 = 0 \) on \( [0, 2\pi) \).
3. Prove that \( \sin^2x + \cos^2x = 1 \) and use it to solve \( 2\cos^2x + 3\sin x - 3 = 0 \).
4. Determine all solutions of \( \tan x = \sqrt{3} \) on \( [0, 2\pi) \).
5. Find all solutions for \( \sin 2x = 0.5 \) on \( [0, 2\pi) \).

**Tip**: Always ensure your calculator or software is set to the correct mode (degrees vs. radians) when solving trigonometric equations!

select all solutions to the equation 2sin^2x + sinx - 1 = 0 on the interval [0, 2pi)

Learn how to solve the trigonometric equation 2sin^2x + sinx - 1 = 0 on the interval [0, 2pi) with a detailed explanation. Covers quadratic equations, trigonometric functions, and sine values from the unit circle. Perfect for Grades 10-12 students.

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