The equation you've provided is:

$\cos(6x) = \sin(2x + 18^\circ)$

To solve this equation, we can use the identity that relates sine and cosine:

$\sin(A) = \cos(90^\circ - A)$

Using this identity, we can rewrite the sine function on the right-hand side of the equation:

$\sin(2x + 18^\circ) = \cos\left(90^\circ - (2x + 18^\circ)\right)$ $= \cos(72^\circ - 2x)$

Now the equation becomes:

$\cos(6x) = \cos(72^\circ - 2x)$

Since the cosine function is periodic, two general conditions arise for equality:

1. $6x = 72^\circ - 2x + 360^\circ n$, where $n \in \mathbb{Z}$ (the general solution).
2. $6x = -(72^\circ - 2x) + 360^\circ n$, also accounting for cosine's periodicity.

Case 1: $6x = 72^\circ - 2x + 360^\circ n$

$6x + 2x = 72^\circ + 360^\circ n$ $8x = 72^\circ + 360^\circ n$ $x = \frac{72^\circ + 360^\circ n}{8}$ $x = 9^\circ + 45^\circ n$

Case 2: $6x = -(72^\circ - 2x) + 360^\circ n$

$6x = -72^\circ + 2x + 360^\circ n$ $6x - 2x = -72^\circ + 360^\circ n$ $4x = -72^\circ + 360^\circ n$ $x = \frac{-72^\circ + 360^\circ n}{4}$ $x = -18^\circ + 90^\circ n$

Final Solutions:

Thus, the general solutions for $x$ are:

1. $x = 9^\circ + 45^\circ n$
2. $x = -18^\circ + 90^\circ n$

where $n$ is an integer.

Would you like to go into more detail, or have any questions?

Here are some related questions:

1. What are the periodic properties of trigonometric functions?
2. How would the solution change if a different trigonometric identity was used?
3. How can these solutions be represented on a unit circle?
4. What happens if you graph both sides of the equation?
5. How would the solution approach change if radians were used instead of degrees?

Tip: Always check the domain and periodicity of trigonometric functions to ensure all solutions are accounted for.