Question

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

Answer 1

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

`x_1 = -pi/3`

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

Explanation:

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`