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

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Answer 1 · 2017-10-10T13:54:56

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

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#