We start by rewriting the given equation:

$-2\sin^2 x = \sin x - 1$

Step 1: Rearrange the equation

Rearrange all terms to one side to set the equation to 0:

$-2\sin^2 x - \sin x + 1 = 0$

Multiply through by $-1$ to simplify:

$2\sin^2 x + \sin x - 1 = 0$

Step 2: Solve as a quadratic equation

Let $u = \sin x$. The equation becomes:

$2u^2 + u - 1 = 0$

Solve this quadratic equation using the quadratic formula:

$u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here:

• $a = 2$,
• $b = 1$,
• $c = -1$.

Substitute these values:

$u = \frac{-1 \pm \sqrt{1^2 - 4(2)(-1)}}{2(2)}$

$u = \frac{-1 \pm \sqrt{1 + 8}}{4}$

$u = \frac{-1 \pm \sqrt{9}}{4}$

$u = \frac{-1 \pm 3}{4}$

So:

$u = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2}, \quad u = \frac{-1 - 3}{4} = \frac{-4}{4} = -1$

Thus, $u = \sin x = \frac{1}{2}$ or $\sin x = -1$.

Step 3: Solve for $x$ in $[0, 2\pi)$

Case 1: $\sin x = \frac{1}{2}$

The solutions for $\sin x = \frac{1}{2}$ are:

$x = \frac{\pi}{6}, \frac{5\pi}{6}$

Case 2: $\sin x = -1$

The solution for $\sin x = -1$ is:

$x = \frac{3\pi}{2}$

Final Answer:

The solutions to the equation in $[0, 2\pi)$ are:

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

Would you like to see a detailed explanation of any specific step?

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1. What are the properties of sine functions useful here?
2. Can you derive solutions for $\sin x = \frac{1}{2}$ using the unit circle?
3. Why is the quadratic substitution $u = \sin x$ effective?
4. How can you verify these solutions graphically or numerically?
5. What does the interval $[0, 2\pi)$ represent for periodic functions?

Tip: To check solutions, substitute them back into the original equation!