Find the exact values of the vertical asymptotes for o is less than or equal to Q is less than or equal to pi for the function y= cot(3Q)

I need to show my work for this one but I do not know how to do this.

To find the vertical asymptotes of a function, you can follow these steps:

1. Determine the values of Q that make the function undefined. In this case, the function is y = cot(3Q).

The cotangent function is undefined whenever the cosine function is equal to 0 because cot(Q) = cos(Q) / sin(Q). So, we need to find the values of Q for which cos(3Q) = 0.

2. Solve the equation cos(3Q) = 0 for Q.

To find the values of Q that make cos(3Q) = 0, you need to find the solutions to the equation 3Q = π/2 + kπ or 3Q = 3π/2 + kπ, where k is an integer. These equations represent the values in which the cosine function crosses the x-axis.

Divide both sides of each equation by 3 to solve for Q:

a. For 3Q = π/2 + kπ:
Q = (π/2 + kπ)/3

b. For 3Q = 3π/2 + kπ:
Q = (3π/2 + kπ)/3

These solutions represent the values at which the cotangent function has vertical asymptotes.

3. Determine the range of Q values within the interval o ≤ Q ≤ π.

Since we are given 0 ≤ Q ≤ π, we need to determine the values of Q within this range that satisfy the above solutions.

For a: (π/2 + kπ)/3
When k = 0, Q = (π/2)/3 = π/6 (within the given range)
When k = 1, Q = (π/2 + π)/3 = π/2 (within the given range)

So, within the range 0 ≤ Q ≤ π, Q = π/6 and Q = π/2 are the values that satisfy the equation 3Q = π/2 + kπ.

For b: (3π/2 + kπ)/3
When k = 0, Q = (3π/2)/3 = π/2 (outside the given range)
When k = 1, Q = (3π/2 + π)/3 = 2π/3 (outside the given range)

So, within the range 0 ≤ Q ≤ π, no values satisfy the equation 3Q = 3π/2 + kπ.

4. Finalize the solution.

The values within the given range that satisfy the equation cos(3Q) = 0 (and consequently make the function undefined) are Q = π/6 and Q = π/2. These values correspond to the vertical asymptotes of the function y = cot(3Q) within the range 0 ≤ Q ≤ π.