Answer:
θ = π/6, π/2, 5π/6, -π/6, -π/2, -5π/6, and so on.
Explanation:
To find the solution set for the equation sin(3θ) = 1, you can use trigonometric properties.
The sine function has a range between -1 and 1, so it can't be equal to 1 for most values of θ. However, it is possible for sin(3θ) to be equal to 1 if θ is such that 3θ is an odd multiple of π/2, as sin(π/2) = 1.
So, you can set up the equation:
3θ = (2n + 1)π/2, where n is an integer.
Now, solve for θ:
θ = ((2n + 1)π/2) / 3
The solution set for θ is all values that can be obtained by plugging in integer values for n into the equation. The set of solutions is:
θ = (π/2)/3, (3π/2)/3, (5π/2)/3, (-π/2)/3, (-3π/2)/3, (-5π/2)/3, and so on, where n is an integer.
So, the solution set is:
θ = π/6, π/2, 5π/6, -π/6, -π/2, -5π/6, and so on.