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
$x=2npi-pi/3+pi/4=2npi-pi/12$

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