Question #1285e

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Answer 1 · 2017-12-30T12:47:01

#x= 2npi+pi/2 and x=2npi" where "n in ZZ#

#(1 + cosx) (1 + sinx) = 2 #

#=>1 + cosx + sinx +sinxcosx= 2 #

#=>cosx + sinx +sinxcosx= 1 #

#=>2sinxcosx+2(sinx+cosx)= 2 #

#=>(sinx+cosx)^2-1+2(sinx+cosx)= 2 #

#=>(sinx+cosx)^2+2(sinx+cosx)=3#

#=>(sinx+cosx)^2+2(sinx+cosx)+1^2=4#

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

#=>(sinx+cosx+3)(sinx+cosx-1)=0#

So #(sinx+cosx+3)=0 #

#=> sinx+cosx=-3-># not possible

when

#(sinx+cosx-1)=0#

#=>cosx+sinx=1#

#=>1/sqrt2cosx+1/sqrt2sinx=1/sqrt2#

#=>cos(pi/4)cosx+sin(pi/4)sinx=cos(pi/4)#

#=>cos(x-pi/4)=cos(pi/4)#

#=>x-pi/4=2npipmpi/4" where "n in ZZ#

Hence #x= 2npi+pi/2 and x=2npi" where "n in ZZ#