How do you find all solutions of #tan(x/2)-sinx=0# in the interval #[0,2pi)#?

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Answer 1 · 2017-11-14T10:53:35

#0 , pi/2 , pi , (3pi)/2#

#tan(x/2)= (sin(x/2))/(cos(x/2))=(sqrt(1/2(1-cosx)))/(sqrt(1/2(1+cosx)))#

#(sqrt(1/2(1-cosx)))/(sqrt(1/2(1+cosx)))-sinx=0#

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

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

#1-cosx=sin^2x+sin^2xcosx#

#1-cosx=1-cos^2x+sin^2xcosx#

#-cosx=-cos^2x+sin^2xcosx#

#cos^2x-sin^2xcosx-cosx=0#

#cosx(1-sin^2x-1)=0#

#cosx(-sin^2x)=0#

#cosx =0=>x=pi/2 , (3pi)/2#

#sin^2x=0=>x=0, pi#