How do you solve #2cosx+2sinx=sqrt2#?

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Answer 1 · 2016-10-27T13:13:00

As below

#2cosx+2sinx=sqrt2#

Dividing both sides by #2sqrt2#

#=>2/(2sqrt2)cosx+2/(2sqrt2)sinx=(sqrt2)/(2sqrt2)#

#=>1/sqrt2cosx+1/sqrt2sinx=1/2#

#=>cosxcos(pi/4)+sinxsin(pi/4)=1/2#

#=>cos(x-pi/4)=cos(pi/3)#

#=>x-pi/4=2npi+-pi/3," where "n in ZZ#

So #x=2npi+pi/3+pi/4=2npi+(7pi)/12#

Or
#x=2npi-pi/3+pi/4=2npi-pi/12#