We are given the trigonometric equation:

$-2 \frac{\cos(x)}{\sin^2(x)} + \frac{\sin(x)}{\cos^2(x)} = 0$

Let's solve this step by step.

Step 1: Combine the terms on the left-hand side

We have two fractions, so let's find a common denominator:

$-2 \frac{\cos(x)}{\sin^2(x)} + \frac{\sin(x)}{\cos^2(x)}$

The common denominator will be $\sin^2(x) \cos^2(x)$. So we rewrite each term:

$-2 \frac{\cos(x)}{\sin^2(x)} = -2 \frac{\cos^3(x)}{\sin^2(x) \cos^2(x)}$

$\frac{\sin(x)}{\cos^2(x)} = \frac{\sin^3(x)}{\sin^2(x) \cos^2(x)}$

Now we combine them:

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

Step 2: Set the numerator equal to zero

For the fraction to be zero, the numerator must be zero:

$-2 \cos^3(x) + \sin^3(x) = 0$

Step 3: Factor the expression

We can factor the left-hand side as a difference of cubes:

$\sin^3(x) - 2 \cos^3(x) = (\sin(x) - \cos(x))( \sin^2(x) + \sin(x) \cos(x) + \cos^2(x))$

Since $\sin^2(x) + \cos^2(x) = 1$, the expression simplifies to:

$(\sin(x) - 2 \cos(x))(1 + \sin(x) \cos(x)) = 0$

Step 4: Solve for $x$

Now, we have two cases to solve:

Case 1: $\sin(x) - 2 \cos(x) = 0$

This can be rewritten as:

$\sin(x) = 2 \cos(x)$

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

$\tan(x) = 2$

So, $x = \tan^{-1}(2) + n\pi$, where $n$ is any integer.

Case 2: $1 + \sin(x) \cos(x) = 0$

We use the identity $\sin(2x) = 2 \sin(x) \cos(x)$, so this becomes:

$1 + \frac{1}{2} \sin(2x) = 0$

$\sin(2x) = -2$

But the sine function has a range of $[-1, 1]$, so there are no solutions for this case.

Final Solution:

The solution to the equation is:

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

Would you like more details or have any questions?

Here are 5 related questions:

1. How can we solve trigonometric equations involving multiple angles?
2. What are common trigonometric identities used in solving equations?
3. How do we handle cases where no solution exists within the range of a trigonometric function?
4. What are some alternative methods for factoring trigonometric expressions?
5. How does the range of inverse trigonometric functions affect the solutions?

Tip: When solving trigonometric equations, always check if the solutions fall within the valid range of the function (e.g., sine and cosine have ranges of $[-1, 1]$).