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 $\displaystyle \alpha = k \pi$ implies that $\displaystyle \alpha = 2\pi$ would be a solution, but plugging in that solution for alpha doesn't work:

$\displaystyle 2 \cos( \alpha /2 ) = \cos \alpha + 1$

$\displaystyle 2 \cos( 2\pi /2 ) = \cos 2\pi + 1$

$\displaystyle 2(-1) = 1 + 1$

$\displaystyle -2 = 2 ????$

(This is what I get for reading too fast.

)

So, Krizalid's method is probably the best one:

Let $\displaystyle \alpha = 2t$, then

$\displaystyle \begin{aligned}

2\cos t &= \cos (2t) + 1 \\

2\cos t &= 2\cos^2(2t) - 1 + 1 \\

2\cos t &= 2\cos^2 t \\

2\cos^2 t - 2\cos t &= 0 \\

2\cos t(cos t - 1) &= 0 \\

\end{aligned}$

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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