Thanks! can you show me the algebraic step for Q2?
The answer is 4kpi...
mine looks like this:
cos(x/2)-1=0
cos(x/2)=1 which is pi +kpi No, This means that x/2 = 2k pi, because cos(even multiples of pi)=1 and cos(oddmultiples of pi) = -1. The 2 in the 2k pi means even multiples of pi So x/2 = 2k pi So x = 4k pi
cos(x/2)= (pi + kpi)2 Once you've brought in your pi's you don't have the cos there anymore.
cosx=2pi+2kpi and you can't say that if cos(x/2) = n then cos x = 2n You just can't take the 2 out like that.
is that right so far?