How do you solve #tanx+sqrt3=0#?

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Answer 1 · 2015-06-16T17:18:55

#tan(x)+sqrt3=0# has two solutions:

#x_1 = -pi/3#

#x_2 = pi-pi/3 = (2pi)/3#

The equation #tan(x)+sqrt3=0# can be rewritten as
#tan(x)=-sqrt3#

Knowing that #tan(x) = sin(x)/cos(x)#

and knowing some specific values of #cos# and #sin# functions:

#cos(0)=1# ; #sin(0)=0#

#cos(pi/6)=sqrt3/2# ; #sin(pi/6)=1/2#

#cos(pi/4)=sqrt2/2# ; #sin(pi/4)=sqrt2/2#

#cos(pi/3)=1/2# ; #sin(pi/3)=sqrt3/2#

#cos(pi/2)=0# ; #sin(pi/2)=1#

as well as the following #cos# and #sin# properties:

#cos(-x)=cos(x)# ; #sin(-x)=-sin(x)#
#cos(x+pi)=-cos(x)# ; #sin(x+pi)=-sin(x)#

We find two solutions:

1) #tan(-pi/3) = sin(-pi/3)/cos(-pi/3) = (-sin(pi/3))/cos(pi/3) = - (sqrt3/2)/(1/2) = -sqrt3#

2) #tan(pi-pi/3) = sin(pi-pi/3)/cos(pi-pi/3) = (-sin(-pi/3))/(-cos(-pi/3)) = sin(pi/3)/(-cos(pi/3)) = - (sqrt3/2)/(1/2) = -sqrt3#