An equation is given. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to three decimal places where appropriate. If there is no solution, enter NO SOLUTION.)− 23 cos(3𝜃) = 1(a) Find all solutions of the equation.𝜃 = (b) Find the solutions in the interval [0, 2𝜋).𝜃 =

(a) To find all solutions of the equation, we first isolate cos(3θ) by dividing both sides of the equation by -23:

cos(3θ) = -1/23

Next, we find the angle whose cosine is -1/23. We do this by taking the inverse cosine (or arccos) of both sides:

3θ = arccos(-1/23)

To get θ by itself, we divide both sides by 3:

θ = arccos(-1/23) / 3

This gives us the general solution for θ. However, because the cosine function has a period of 2π, we can add 2πk (where k is any integer) to this solution to get all possible solutions:

θ = arccos(-1/23) / 3 + 2πk

(b) To find the solutions in the interval [0, 2π), we substitute integer values for k into the general solution until we get values of θ that fall within this interval.

For example, if k=0, we get:

θ = arccos(-1/23) / 3

If this value is within the interval [0, 2π), it is a solution. We continue this process with k=1, k=2, etc., until we get values of θ that are outside the interval.

Note: The actual numerical solutions will depend on the value of arccos(-1/23), which can be calculated using a calculator.