We are tasked with solving the equation:

\[
\tan^2(x) - (\sqrt{3} - 1)\tan(x) = 0
\]

### Step 1: Factor the equation
This is a quadratic equation in terms of \(\tan(x)\), so we can factor it as follows:

\[
\tan(x) \left( \tan(x) - (\sqrt{3} - 1) \right) = 0
\]

### Step 2: Solve for \(\tan(x)\)
Now, we set each factor to zero and solve for \(\tan(x)\).

#### Case 1: 
\[
\tan(x) = 0
\]

This gives the solutions:
\[
x = n\pi, \quad n \in \mathbb{Z}
\]
(where \(n\) is any integer, as the tangent function has a period of \(\pi\)).

#### Case 2: 
\[
\tan(x) = \sqrt{3} - 1
\]

To find the solutions for this, we solve:
\[
x = \tan^{-1}(\sqrt{3} - 1)
\]
Using a calculator or approximation, we get:
\[
x \approx \tan^{-1}(0.732) \approx 0.634 \text{ radians} + n\pi, \quad n \in \mathbb{Z}
\]
(as the general solution for tangent is also periodic with \(\pi\)).

### Final solution:
\[
x = n\pi \quad \text{or} \quad x \approx 0.634 + n\pi, \quad n \in \mathbb{Z}
\]

Would you like any details or further clarifications?

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Here are 5 related questions you might explore:
1. How can we solve other trigonometric equations involving tangent?
2. What is the general solution for \(\tan(x) = k\)?
3. How do the periodic properties of trigonometric functions affect their solutions?
4. Can you solve a similar equation involving sine or cosine?
5. How does factoring help in solving quadratic equations?

**Tip:** Always remember that trigonometric functions are periodic, so when solving equations like this, include the general form of the solution with \(\pm n\pi\).

Explore how to solve the trigonometric quadratic equation tan^2(x) - (√3 - 1)tan(x) = 0 using factorization and trigonometric identities. This detailed solution is ideal for students in Grades 10-12, covering key concepts in trigonometry and quadratic equations.

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