How do you solve sin(x/2)=1-cosx?

1 Answer
Oct 10, 2017

x=2kpi or x=pi/3+2kpi or x=5/3pi+2kpi

Explanation:

We can write that

cosx=cos^2(x/2)-sin^2(x/2)

and

1=cos^2(x/2)+sin^2(x/2)

Then let's substitute:

sin(x/2)=cancel(cos^2(x/2))+sin^2(x/2)cancel(-cos^2(x/2))+sin^2(x/2)

Let's sum and get the equivalent equation:

2sin^2(x/2)-sin(x/2)=0

that becomes:

sin(x/2)(2sin(x/2)-1)=0

It is solved by:

1) sin(x/2)=0->x/2=kpi->x=2kpi

2) 2sin(x/2)-1=0->sin(x/2)=1/2->x/2=pi/6+2kpi or x/2=5/6pi+2kpi->x=pi/3+2kpi or x=5/3pi+2kpi