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.

Learn how to solve the trigonometric equation sin(x) = cos(π/6). This problem covers core trigonometric concepts, such as sine, cosine functions, and their general solutions, and is ideal for high school students in Grades 10-12.

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