How do you solve #3\sin x + \sqrt { 3} \cos x = 0#?

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Answer 1 · 2017-02-11T10:08:21

#x=npi-pi/6#, where #n# is an integer.

Dividing each side in the equation #3sinx+sqrt3cosx=0# by #2sqrt3#.

This number has been obtained by adding the squares of the coefficients and taking their square root i.e.

#sqrt(3^2+(sqrt3)^2)=sqrt12=2sqrt3# and we get

#sqrt3/2sinx+1/2cosx=0#

or #sinx xxcos(pi/6)+cosx xxsin(pi/6)=0#

or #sin(x+pi/6)=0#

But as for every #npi#, where #n# is an integer, we have sin(npi)=0#, we have

#x+pi/6=npi#

or #x=npi-pi/6#, where #n# is an integer.