From you get , from which you have the solutions
The only way you could get three solutions would be to have an equation with or in it, for some . Also, the solutions would be equally spaced, which your given solution is not.
Was hoping I could get some direction in this one, can't seem to get the desired results.
Solve for all possible
Now, using the half angle identity I get
This is where I'm at. The only thing I can think of is to factor out a
but then I'm left with
The answers to the problem are:
So I'm pretty sure I'm missing something here.
Now that I looked at the problem more closely, the problem with your method, which icemanfan and I expanded on...
(This is what I get for reading too fast. )
So, Krizalid's method is probably the best one:
Let , then
2 cos t = 0 or cos t = 0, which means that
t = π/2 + 2πk and t = 3π/2 + 2πk
But we said α = 2t, so
α = π + 4πk and α = 3π + 4πk
[or α = 180° + (720°)k or α = 540° + (720°)k].
cos t - 1 = 0 or cos = 1, which means that
t = 0 + 2πk or t = 2πk
But again, α = 2t, so
α = 4πk
[or α = (720°)k].