General solution for #2sin^2x+cosx=1#?

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Answer 1 · 2018-06-05T15:07:07

#x={2kpi+-(2pi)/3,kinZZ}uu{2kpi,kinZZ}#

Here,

#2sin^2x+cosx=1#

#=>2(1-cos^2x)+cosx=1#

#=>2-2cos^2x+cosx=1#

#=>2cos^2x-cosx-1=0#

#=>2cos^2x-2cosx+cosx-1=0#

#=>2cosx(cosx-1)+1(cosx-1)=0#

#=>(2cosx+1)(cosx-1)=0#

#=>2cosx=-1 or cosx=1#

#=>cosx=-1/2 or cosx=1#

#(i)cosx=-1/2=cos(pi-pi/3)#

#=>cosx=cos((2pi)/3)#

#=>x=2kpi+-(2pi)/3,kinZZ#

#(ii)cosx=1=>x=2kpi,kinZZ#

Hence, the general solution is :

#x={2kpi+-(2pi)/3,kinZZ}uu{2kpi,kinZZ}#