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The equation you provided is:

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

We are also given that one solution is \( x = \frac{\pi}{2} \). Let's solve this equation step by step.

### Step 1: Let \( u = \sin(x) \)
This transforms the equation into a quadratic form:
\[
2u^2 - u - 1 = 0
\]

### Step 2: Solve the quadratic equation
We can use the quadratic formula:
\[
u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 2 \), \( b = -1 \), and \( c = -1 \). Plugging in the values:

\[
u = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-1)}}{2(2)}
\]
\[
u = \frac{1 \pm \sqrt{1 + 8}}{4}
\]
\[
u = \frac{1 \pm \sqrt{9}}{4}
\]
\[
u = \frac{1 \pm 3}{4}
\]

This gives two solutions for \( u \):
\[
u = \frac{1 + 3}{4} = 1 \quad \text{or} \quad u = \frac{1 - 3}{4} = -\frac{1}{2}
\]

### Step 3: Solve for \( x \)
Now, recall that \( u = \sin(x) \), so we have two cases:

#### Case 1: \( \sin(x) = 1 \)
The solution for \( \sin(x) = 1 \) is:
\[
x = \frac{\pi}{2} + 2n\pi \quad \text{for} \quad n \in \mathbb{Z}
\]

#### Case 2: \( \sin(x) = -\frac{1}{2} \)
The general solutions for \( \sin(x) = -\frac{1}{2} \) are:
\[
x = \frac{7\pi}{6} + 2n\pi \quad \text{or} \quad x = \frac{11\pi}{6} + 2n\pi \quad \text{for} \quad n \in \mathbb{Z}
\]

### Final Solutions:
The solutions to the equation are:
\[
x = \frac{\pi}{2} + 2n\pi, \quad x = \frac{7\pi}{6} + 2n\pi, \quad x = \frac{11\pi}{6} + 2n\pi
\]
where \( n \) is any integer.

Would you like more details or have any questions?

Here are 5 related questions to expand on this topic:
1. How can we derive the general solution for trigonometric equations?
2. What are the different methods for solving trigonometric equations like this one?
3. How does the unit circle help in solving \( \sin(x) = \text{constant} \)?
4. What happens when we solve the same equation for cosine instead of sine?
5. How do we find all possible solutions within a specific interval?

**Tip:** Always check for extraneous solutions when dealing with trigonometric equations, especially if additional conditions are given (like an interval).

Solve the trigonometric equation 2sin^2(x) - sin(x) - 1 = 0 for x, given that one solution is x = π/2.

Learn how to solve the trigonometric equation 2sin²(x) - sin(x) - 1 = 0 step by step, including using the quadratic formula and solving for sine. This problem covers key concepts like solving quadratic equations in trigonometry and finding the general solutions for sine. Suitable for high school students.

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