Why there are two different values of θ for same quadrant?

Question

Let Sin θ = 1/2 is function. Let us find its solution set.

sine is +ve in I and II quadrant with reference angle π/6

θ = π/6 (I quadrant)

Now here is my problem.

We can use π-θ and (π/2)+θ to find values in 2nd quadrant.

But when we use first case (π-θ)

θ = π-π/6 = 5π/6 = 150 degree

And when we use 2nd case [(π/2)+θ]

θ = (π/2)+π/6 = 8π/12 = 120 degree

Now these values are different. Please help why are they different.

Also please tell me in clock wise which of following we take:

(π/2)-θ , π-θ , (3π/2)-θ , 2π-θ

or

(π/2)-θ ,(π/2)+θ , π+θ , (3π/2)+θ

Answer 1

There aren't two values of theta for the same quadrant. $\sin \frac {\pi}{6}=\frac{1}{2}$ and $\sin \frac {5\pi}{6}=\frac{1}{2}$, that's it. Those two angles have a sine (y/r) of 1/2.