Let's solve the given equation step by step:

$9\sin(18x) + 11 = 2$

First, isolate the sine term:

$9\sin(18x) = 2 - 11$ $9\sin(18x) = -9$

Now, divide both sides by 9:

$\sin(18x) = -1$

The general solution for $\sin(\theta) = -1$ is:

$\theta = \dfrac{3\pi}{2} + 2n\pi \quad \text{where } n \text{ is any integer}$

Substitute $\theta = 18x$:

$18x = \dfrac{3\pi}{2} + 2n\pi$

Now, solve for $x$:

$x = \dfrac{3\pi}{36} + n\cdot\dfrac{\pi}{9}$

Simplify the fraction:

$x = \dfrac{\pi}{12} + n\cdot\dfrac{\pi}{9}$

This is one possible solution. However, let's consider the additional periodic solutions.

The sine function $\sin(18x)$ has a period of $\dfrac{2\pi}{18} = \dfrac{\pi}{9}$, so other solutions within the period might exist.

Considering the periodicity and the general solution, the correct forms must match the equivalent angles where $\sin(18x) = -1$ at $\dfrac{3\pi}{2}$ and its periodic multiples.

So let's check the options:

• Choice A: $-\dfrac{\pi}{2} + n\cdot 2\pi$ does not match our derived solution.
• Choice B: $-\dfrac{\pi}{36} + n\cdot\dfrac{\pi}{9}$ does not match the correct angles either.
• Choice C: $n\cdot\pi$ is not correct, as the sine function only equals $-1$ at specific angles, not all multiples of $\pi$.
• Choice D: $\dfrac{\pi}{36} + n\cdot\dfrac{\pi}{9}$ does not match.
• Choice E: $\dfrac{\pi}{2} + n\cdot\dfrac{\pi}{11}$ is incorrect because $\dfrac{\pi}{11}$ is unrelated to the actual period of the function.
• Choice F: $\dfrac{\pi}{2} + n\cdot\pi$ is unrelated to the solution derived.

None of the options provided accurately represent the solution derived from $\sin(18x) = -1$, which is:

$x = \dfrac{\pi}{12} + n\cdot\dfrac{\pi}{9}$

Therefore, none of the answer choices correspond to the correct solution set.

Would you like further clarification on any part of the solution?

Here are some related questions:

1. What would be the general solution for $\sin(kx) = 1$?
2. How does the periodicity of the sine function affect the number of solutions within a given interval?
3. How can we use the unit circle to determine the exact angles where $\sin(x) = -1$?
4. What if the equation involved $\cos(18x) = -1$ instead of sine—how would the solutions change?
5. How do different coefficients of $x$ inside a sine function influence the solution set?

Tip: Always double-check the periodicity of the trigonometric function when solving equations to ensure all possible solutions are considered.