Let's solve the equation:

\[
-2 \cos^3(x) + \frac{\sin^3(x)}{\sin^2(x)} \cdot \cos^2(x) = 0
\]

### Step 1: Simplify the equation
We can simplify the second term. The fraction \(\frac{\sin^3(x)}{\sin^2(x)}\) simplifies to:

\[
\frac{\sin^3(x)}{\sin^2(x)} = \sin(x)
\]

So the equation becomes:

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

### Step 2: Factor the equation
Now, factor out \(\cos^2(x)\):

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

### Step 3: Solve for \(\cos^2(x) = 0\)
Since this is a product, either \(\cos^2(x) = 0\) or \(-2 \cos(x) + \sin(x) = 0\).

For \(\cos^2(x) = 0\), we know that \(\cos(x) = 0\). The solutions to \(\cos(x) = 0\) are:

\[
x = \frac{\pi}{2} + n\pi \quad \text{where} \quad n \in \mathbb{Z}
\]

### Step 4: Solve for \(-2 \cos(x) + \sin(x) = 0\)
Now solve the second factor:

\[
-2 \cos(x) + \sin(x) = 0
\]

This can be rearranged as:

\[
\sin(x) = 2 \cos(x)
\]

Dividing both sides by \(\cos(x)\) (assuming \(\cos(x) \neq 0\)):

\[
\tan(x) = 2
\]

The general solution to \(\tan(x) = 2\) is:

\[
x = \tan^{-1}(2) + n\pi \quad \text{where} \quad n \in \mathbb{Z}
\]

### Final Solutions:
1. \(x = \frac{\pi}{2} + n\pi\), where \(n \in \mathbb{Z}\)
2. \(x = \tan^{-1}(2) + n\pi\), where \(n \in \mathbb{Z}\)

Would you like further details or have any questions?

Here are five related questions to deepen your understanding:
1. How can you graph the functions involved in this solution?
2. What are the steps for solving other trigonometric equations?
3. How can you solve \(\sin(x) = \cos(x)\)?
4. What is the range of the arctangent function?
5. Can you derive a similar equation using \(\tan(x)\) from the start?

**Tip**: When solving trigonometric equations, it's helpful to use identities or factorization to simplify complex terms!

-2cos^3(x) + sin^3(x)/sin^2(x) * cos^2(x) = 0

Learn how to solve the trigonometric equation -2cos^3(x) + sin^3(x)/sin^2(x) * cos^2(x) = 0 using trigonometric identities, factorization, and general solutions for cos(x) and tan(x). This problem covers advanced trigonometric concepts and is suitable for students in Grades 10-12.

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