Solve sin(3θ) = -√2/2 for θ in the interval [0,2π) for the first four solutions only.

Question

Solve sin(3θ) = -√2/2 for θ in the interval [0,2π) for the first four solutions only.

Enter your answer in exact form and in ascending order.

Answer 1

sin(3θ) = -√2/2

Recall that sin(π/4) = √2/2

Since sine is negative, 3θ lies in quadrant 3 or 4.

So, 3θ = 5π/4 + 2kπ or 7π/4 + 2kπ

θ = 5π/12 + 2kπ/3 or 7π/12 + 2kπ/3, where k = 0, ±1, ±2, ...

If k = 0, we get θ = 5π/12 or 7/12

If k = 1, we have θ = 13π/12 or 15π/12

Answer 2

sine is negative in quadrants 3 and 4:

sin T = sqrt(2)/2 means that the BASE angle is 45 degrees

so the first two solutions are 180+45 = 225 = 225/180 * pi = 45/36 * pi = 5*pi/4

and

360-45 =315 = 315/180 * pi = 63/36 * pi = 7*pi/4

Add 360 = 2 *pi to these to get your next two solutions