Use "quadratic" factoring to solve the following equation for all three answers on the interval 0≤0 ≤ 2π
2 sin2 x - 3 sin x + 1 = 0

The given equation is:

\[
2 \sin^2 x - 3 \sin x + 1 = 0
\]

We can solve this equation using factoring by treating it as a quadratic equation in terms of \(\sin x\). Let \( u = \sin x \), which transforms the equation into:

\[
2u^2 - 3u + 1 = 0
\]

### Step 1: Factor the quadratic equation
We need to factor \( 2u^2 - 3u + 1 \).

Looking for factors of \(2 \cdot 1 = 2\) that add up to \(-3\), we find that the factors are \(-1\) and \(-2\). Therefore, we rewrite the middle term:

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

Now, group the terms:

\[
u(2u - 1) - 1(2u - 1) = 0
\]

Factoring gives us:

\[
(2u - 1)(u - 1) = 0
\]

### Step 2: Solve for \(u = \sin x\)
Now, set each factor equal to zero:

1. \(2u - 1 = 0 \Rightarrow u = \frac{1}{2}\)
2. \(u - 1 = 0 \Rightarrow u = 1\)

Thus, we have two cases:
- \(\sin x = \frac{1}{2}\)
- \(\sin x = 1\)

### Step 3: Solve for \(x\) in the interval \(0 \leq x \leq 2\pi\)

1. For \(\sin x = \frac{1}{2}\):
   - The general solutions for \(\sin x = \frac{1}{2}\) are \(x = \frac{\pi}{6}\) and \(x = 5\pi/6\) (from the unit circle).

2. For \(\sin x = 1\):
   - The general solution for \(\sin x = 1\) is \(x = \frac{\pi}{2}\).

### Step 4: Final solutions
The solutions to the equation \(2 \sin^2 x - 3 \sin x + 1 = 0\) in the interval \(0 \leq x \leq 2\pi\) are:

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

Would you like further details on this solution, or do you have any questions?

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Here are 5 related questions for practice:
1. Solve \( \sin^2 x - 5 \sin x + 6 = 0 \) for \( 0 \leq x \leq 2\pi \).
2. What are the general solutions to \( \sin x = \frac{\sqrt{3}}{2} \)?
3. How do you solve \( 3 \cos^2 x - 2 \cos x - 1 = 0 \) using factoring?
4. Find the solutions of \( 2 \tan^2 x - 3 \tan x + 1 = 0 \).
5. Solve \( 4 \sin^2 x - 4 \sin x + 1 = 0 \) in the interval \( 0 \leq x \leq 2\pi \).

**Tip:** Always verify your solutions by substituting them back into the original trigonometric equation to ensure they satisfy the equation.

Use 'quadratic' factoring to solve the following equation for all three answers on the interval 0≤ x ≤ 2π: 2 sin²(x) - 3 sin(x) + 1 = 0

Learn how to solve the trigonometric equation 2 sin²(x) - 3 sin(x) + 1 = 0 on the interval 0 ≤ x ≤ 2π using quadratic factoring. This problem covers key concepts from trigonometry and quadratic equations, including solving using the unit circle. Suitable for students in Grades 10-12.

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