Now that I looked at the problem more closely, the problem with your method, which icemanfan and I expanded on...
... is that by squaring both sides you introduced an extraneous roots. Saying that the solution is
implies that
would be a solution, but plugging in that solution for alpha doesn't work:
(This is what I get for reading too fast.
)
So, Krizalid's method is probably the best one:
Let
, then
So
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].
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