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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