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We are tasked with solving the equation:

\[
2 \tan(x) \sin(x) - 2 \tan(x) = 0
\]

in the interval \([0, 2\pi)\). Let's break it down step by step:

### Step 1: Factor the equation
We can factor out \(2 \tan(x)\) from both terms:

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

Now, we have two possible cases to solve.

### Step 2: Solve each factor
#### Case 1: \(2 \tan(x) = 0\)
This simplifies to:
\[
\tan(x) = 0
\]

We know that \(\tan(x) = 0\) at \(x = 0, \pi, 2\pi\), but since we're in the interval \([0, 2\pi)\), the valid solutions are:
\[
x = 0, \pi
\]

#### Case 2: \(\sin(x) - 1 = 0\)
This simplifies to:
\[
\sin(x) = 1
\]

We know that \(\sin(x) = 1\) at \(x = \frac{\pi}{2}\) in the interval \([0, 2\pi)\).

### Step 3: Combine the solutions
The solutions from both cases are:
\[
x = 0, \pi, \frac{\pi}{2}
\]

Thus, the solutions to the equation \(2 \tan(x) \sin(x) - 2 \tan(x) = 0\) in the interval \([0, 2\pi)\) are:

\[
x = 0, \frac{\pi}{2}, \pi
\]

Would you like a more detailed explanation of any step? 

Here are five related questions to deepen your understanding:
1. How do you determine the general solutions for \(\tan(x) = 0\)?
2. Why is factoring an effective method in solving trigonometric equations?
3. What are the other values of \(\tan(x) = 0\) beyond the given interval?
4. How does the periodicity of sine and tangent functions affect the solution of this equation?
5. Can this method be used for equations involving other trigonometric identities?

**Tip**: When solving trigonometric equations, always consider the fundamental period of the functions involved to capture all solutions within the given interval.

Solve 2 tan(x) sin(x) - 2 tan(x) = 0 in the interval [0, 2π).

Learn how to solve the trigonometric equation 2 tan(x) sin(x) - 2 tan(x) = 0 in the interval [0, 2π). This step-by-step solution involves key concepts such as factoring, trigonometric identities, and the zero product property. Suitable for students in Grades 10-12.

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