How do you solve #sin(x/2) + cosx =0# in the interval [0, 2pi]?

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Answer 1 · 2016-02-07T00:41:26

We know that

#cosx=1-2sin^2(x/2)#

Hence

#sin(x/2)+cos(x)=0=>sin(x/2)+1-2sin^2(x/2)=0=> 2sin^2(x/2)-sin(x/2)-1=0#

Set #u=sin(x/2)# so we have

#2u^2-u-1=0=>u^2-u+u^2-1=0=>u(u-1)+(u-1)(u+1)=0=> (u-1)(2u+1)=0#

For #u=1# we get #sin(x/2)=1=>x/2=2kpi+pi/2=>x=4kpi+pi#

where #k# integer.

And for #u=-1/2# we get #sin(x/2)=-1/2=>x/2=2kpi-pi/6=>x=4kpi-pi/3 or x=4kpi+7pi/3#

But since our solutions belong to #[0,2pi]# the acceptable values are

#x=pi#