Solve `tan(x)+sqrt(3)=0 ` :

`tan(x)+sqrt(3)=0 `

`tan(x)=-sqrt(3) `

`x="Tan"^(-1)(-sqrt(3))`

The tangent has value `sqrt(3) ` when the sine has value `sqrt(3)/2 ` and the cosine has value `1/2 ` . (`tanx=sinx/cosx ` )

The tangent is positive in the first and third quadrants, and negative in the second and fourth quadrants.

Thus `x=-pi/3+npi ` for ` ``n in ZZ ` (n an integer.)

The graph of tan(x) and the line `y=-sqrt(3) ` :

Fist simplify the equation, like so:

tan x = -3^(1/2)

tan x as a rule is equal to sinx / cosx. From there you work backwards to determine what values of sin and cos would equal to negative square root of 3. If the results of sin were square root of 3 /2 and cos was 1/2, then the 2s will cancel, leaving you with square root of 3. Now to determine which angle on the unit circle will give you a negative value. This can happen in the 2nd or 4th quadrant at x = 120 degrees or 300 degrees, or 2pi/3 and 5pi/3 in radians.

Given

`Tan(x)+sqrt(3)=0`

`Tan(x)=-sqrt(3)`

`Tan(x)=-Tan(pi/3)`

``But we know

`-Tan(theta)=Tan(pi/2+theta)`

Therefore

`Tan(x)=Tan(pi/2+pi/3)`

`Tan(x)=Tan((5pi)/6)`

Thus

`x=n pi+(5pi)/6`

where n is an integer.

You will need to memorize the unit circle to solve similar problems. The +n(pi) part of the solution arises from the fact that trigonometric functions repeat themselves in periodic cycles.