I need to find the angle alpha for cos(2a) = -√2 /2 . Can anyone help me with this?

 cos(2a) = -√2 /2

2a must be in the 2nd or 3rd quadrant

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The first thing that I notice is that

  $$\frac{\sqrt2}{2}\\\\
=\frac{\sqrt2\times\sqrt2}{2\sqrt2}\\\\
=\frac{2}{2\sqrt2}\\\\
=\frac{1}{\sqrt2}\\\\\\
so \\\\
cos(2a)=\frac{1}{\sqrt2}\\\\
$Now you should recognise that $ \\\\
cos45^0=\frac{1}{\sqrt2}\\\\$$

So 

2a=180-45, 180+45, 180-45+360, 180+45+360 degrees

2a=135, 225, 495, 585 degrees

a=67°30', 112°30', 247°30', 292°30'

That is all the answers  for  $$0\le a\le360^0$$

I know i have used pre-knowledge that you may not have - if you want me to explain differently/more then let me know.