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: $2y^2 + y - 1 = 0$

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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!