To solve the equation:

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

### Step 1: Substitution
Let:
\[
y = \sin(3x)
\]

This transforms the equation into a quadratic form:
\[
2y^2 - 3y - 2 = 0
\]

### Step 2: Solve the quadratic equation
We can solve this quadratic equation using the quadratic formula:
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
For \( 2y^2 - 3y - 2 = 0 \), \(a = 2\), \(b = -3\), and \(c = -2\).

\[
y = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-2)}}{2(2)}
\]
\[
y = \frac{3 \pm \sqrt{9 + 16}}{4}
\]
\[
y = \frac{3 \pm \sqrt{25}}{4}
\]
\[
y = \frac{3 \pm 5}{4}
\]

So, we have two possible solutions for \(y\):
1. \( y = \frac{3 + 5}{4} = \frac{8}{4} = 2 \)
2. \( y = \frac{3 - 5}{4} = \frac{-2}{4} = -\frac{1}{2} \)

### Step 3: Solve for \( \sin(3x) \)
Now, recall that \( y = \sin(3x) \). So we have two cases to solve:

#### Case 1: \( \sin(3x) = 2 \)
This is impossible since the sine function always lies between \(-1\) and \(1\), i.e., \(-1 \leq \sin(3x) \leq 1\). Therefore, there are no solutions in this case.

#### Case 2: \( \sin(3x) = -\frac{1}{2} \)
Now we solve for \( 3x \) in this case:
\[
\sin(3x) = -\frac{1}{2}
\]
This gives the general solution:
\[
3x = \arcsin\left(-\frac{1}{2}\right) + 2n\pi \quad \text{or} \quad 3x = \pi - \arcsin\left(-\frac{1}{2}\right) + 2n\pi
\]
Since \( \arcsin\left(-\frac{1}{2}\right) = -\frac{\pi}{6} \), we have:

1. \( 3x = -\frac{\pi}{6} + 2n\pi \)
2. \( 3x = \pi + \frac{\pi}{6} + 2n\pi = \frac{7\pi}{6} + 2n\pi \)

### Step 4: Solve for \(x\)
For both cases, divide by 3 to solve for \(x\):

1. \( x = \frac{-\frac{\pi}{6} + 2n\pi}{3} = -\frac{\pi}{18} + \frac{2n\pi}{3} \)
2. \( x = \frac{\frac{7\pi}{6} + 2n\pi}{3} = \frac{7\pi}{18} + \frac{2n\pi}{3} \)

Thus, the solutions for \(x\) are:
\[
x = -\frac{\pi}{18} + \frac{2n\pi}{3} \quad \text{or} \quad x = \frac{7\pi}{18} + \frac{2n\pi}{3}
\]
where \(n\) is any integer.

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Would you like more details on any part of the solution? Here are five related questions you might want to explore:

1. How do you derive the general solution for the sine function?
2. What is the significance of the sine function's range \([-1, 1]\)?
3. Can this type of equation involve other trigonometric functions?
4. How can you apply this method to a similar cosine equation?
5. What are some real-world applications of solving trigonometric equations?

**Tip:** Always verify the domain restrictions of trigonometric functions when solving equations to rule out impossible solutions.

Learn how to solve the trigonometric quadratic equation 2sin^2(3x) - 3sin(3x) - 2 using the quadratic formula and trigonometric identities. This problem covers key concepts in trigonometry and algebra, including the sine function's range and quadratic equations. Suitable for students in Grades 10-12.

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