The given equation is:

\[
2 (\sin(x))^2 - 7 \sin(x) + 3 = 0
\]

This is a quadratic equation in terms of \( \sin(x) \). To solve it, we can apply the quadratic formula or factor it. Let's factor it step by step.

### Step 1: Substitution
Let \( y = \sin(x) \). Then the equation becomes:

\[
2y^2 - 7y + 3 = 0
\]

### Step 2: Factoring the quadratic equation
We need to factor \( 2y^2 - 7y + 3 = 0 \). This can be factored as:

\[
(2y - 1)(y - 3) = 0
\]

### Step 3: Solve for \( y \)
Now, solve each factor:

1. \( 2y - 1 = 0 \implies y = \frac{1}{2} \)
2. \( y - 3 = 0 \implies y = 3 \)

### Step 4: Back-substitute \( y = \sin(x) \)
Now, return to \( \sin(x) \):

1. \( \sin(x) = \frac{1}{2} \)
2. \( \sin(x) = 3 \) (which is not possible because \( \sin(x) \) is always between -1 and 1)

So, only \( \sin(x) = \frac{1}{2} \) is valid.

### Step 5: Solve for \( x \)
The general solution for \( \sin(x) = \frac{1}{2} \) is:

\[
x = \frac{\pi}{6} + 2n\pi \quad \text{or} \quad x = \frac{5\pi}{6} + 2n\pi, \quad n \in \mathbb{Z}
\]

Thus, the solution to the equation is:

\[
x = \frac{\pi}{6} + 2n\pi \quad \text{or} \quad x = \frac{5\pi}{6} + 2n\pi, \quad n \in \mathbb{Z}
\]

Would you like more details or have any questions?

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Here are 5 related questions:
1. How do you solve trigonometric equations involving multiple angles?
2. What are the steps for solving quadratic equations using the quadratic formula?
3. Can sine or cosine ever be greater than 1 or less than -1?
4. How do you find all solutions to a trigonometric equation?
5. What are the key properties of the sine function?

**Tip:** Always check the domain of trigonometric functions when solving equations to avoid invalid solutions.

Learn how to solve the trigonometric quadratic equation 2(sin(x))² - 7sin(x) + 3 = 0 with a step-by-step approach. This solution uses the quadratic formula and covers the properties of the sine function, suitable for students in Grades 9-12.

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