The equation you provided is:

$\sin(x) = \cos\left(\frac{\pi}{6}\right)$

First, we know that:

$\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$

So the equation becomes:

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

Now, we need to solve for $x$. The general solution for $\sin(x) = \frac{\sqrt{3}}{2}$ occurs when:

$x = \frac{\pi}{3} + 2k\pi \quad \text{or} \quad x = \pi - \frac{\pi}{3} + 2k\pi$

Simplifying the second solution:

$x = \frac{2\pi}{3} + 2k\pi$

where $k \in \mathbb{Z}$ (an integer).

Thus, the general solutions for $x$ are:

$x = \frac{\pi}{3} + 2k\pi \quad \text{or} \quad x = \frac{2\pi}{3} + 2k\pi$

Would you like further details or clarification on any part of this? Here are some related questions you might consider:

1. How do you derive the general solution for trigonometric equations?
2. Can we use inverse trigonometric functions to solve this equation?
3. How do you solve $\sin(x) = -\frac{\sqrt{3}}{2}$?
4. What is the difference between the sine and cosine functions graphically?
5. How do you solve trigonometric equations in degrees?

Tip: When solving trigonometric equations, always remember to account for all possible solutions in the given domain.