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The given equation is:

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

We will solve this equation step by step in the interval \( [0, 2\pi) \).

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

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

### Step 2: Solving the quadratic equation
We solve the quadratic equation using the quadratic formula:

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

So the two solutions for \( y \) are:

\[
y = \frac{3 + 5}{4} = 2 \quad \text{and} \quad y = \frac{3 - 5}{4} = -\frac{1}{2}
\]

### Step 3: Back-substitute \( y = \sec(x) \)
Now we solve for \( x \).

1. For \( \sec(x) = 2 \):
   \[
   \sec(x) = \frac{1}{\cos(x)} = 2 \quad \Rightarrow \quad \cos(x) = \frac{1}{2}
   \]

   The solutions for \( \cos(x) = \frac{1}{2} \) in the interval \( [0, 2\pi) \) are:
   \[
   x = \frac{\pi}{3}, \frac{5\pi}{3}
   \]

2. For \( \sec(x) = -\frac{1}{2} \):
   \[
   \sec(x) = \frac{1}{\cos(x)} = -\frac{1}{2} \quad \Rightarrow \quad \cos(x) = -2
   \]

   However, since \( \cos(x) \) must lie between -1 and 1, there are no solutions for \( \cos(x) = -2 \).

### Final Answer:
The only solutions are:

\[
x = \frac{\pi}{3}, \frac{5\pi}{3}
\]

You should enter these two solutions separated by a comma: \( \frac{\pi}{3}, \frac{5\pi}{3} \).

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

Here are some related questions:
1. How does the quadratic formula work in general?
2. What is the domain of the secant function?
3. How do you find the general solution for trigonometric equations?
4. What are some common errors when solving trigonometric equations?
5. How does the cosine function behave over different quadrants?

**Tip**: Always check the range of trigonometric functions, especially when using secant, cosecant, and tangent, to ensure solutions are valid.

Find all solutions of the equation in the interval [0, 2π). 2 sec²(x) - 3 sec(x) - 2 = 0

Learn how to solve the trigonometric equation 2 sec²(x) - 3 sec(x) - 2 = 0 in the interval [0, 2π). This solution uses substitution, the quadratic formula, and trigonometric identities. Suitable for high school students in Grades 10-12.

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