We are given the equation:

\[
\sin(2x) = \sqrt{3} \cdot \cos(2x)
\]

To solve for \( x \), we can divide both sides by \( \cos(2x) \) (assuming \( \cos(2x) \neq 0 \)):

\[
\tan(2x) = \sqrt{3}
\]

Now, solving for \( 2x \):

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

The principal value of \( \tan^{-1}(\sqrt{3}) \) is \( \frac{\pi}{3} \) (or 60 degrees). Therefore:

\[
2x = \frac{\pi}{3} + n\pi \quad \text{(where \( n \) is any integer, due to the periodicity of tangent)}
\]

Dividing by 2 to solve for \( x \):

\[
x = \frac{\pi}{6} + \frac{n\pi}{2}
\]

Thus, the general solution for \( x \) is:

\[
x = \frac{\pi}{6} + \frac{n\pi}{2}, \quad \text{where \( n \) is any integer}.
\]

Would you like more details or have any further questions?

Here are 5 related questions:
1. How do we solve for \( x \) in trigonometric equations involving sine and cosine?
2. What is the general form of the tangent function's solutions?
3. What is the periodicity of the tangent function and how does it affect solutions?
4. Can we apply the same process to solve equations with other trigonometric identities?
5. How do we handle cases where \( \cos(2x) = 0 \)?

**Tip:** Always remember that trigonometric functions like sine and cosine are periodic, meaning their solutions often involve adding multiples of the period to the principal solution.

Learn how to solve the trigonometric equation sin(2x) = sqrt(3) * cos(2x) using the tangent function. This problem involves key trigonometric concepts such as sine, cosine, tangent, and inverse functions. Suitable for students in Grades 10-12.

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