Question

How do you solve `tanx=sqrt3`?

Answer 1

`x = pi/3 + n pi" "` for any integer `n`

Explanation:

Consider a triangle with sides `1`, `sqrt(3)/2` and `2`.

This is a right angled triangle and one half of an equilateral triangle...

Now `tan theta = "opposite"/"adjacent"`

So looking at our diagram, `tan (pi/3) = sqrt(3)/1 = sqrt(3)`

So one solution of the given equation is `x = pi/3`

Note that:

`tan(theta + pi) = sin(theta + pi)/cos(theta + pi) = (-sin(theta))/(-cos(theta)) = sin(theta)/cos(theta) = tan (theta)`

Also note that `tan(theta)` is strictly monotonically increasing and therefore one to one for `theta` in the range `(-pi/2, pi/2)`.

So `tan(theta)` is periodic with period `pi`

Hence we find:

`tan(pi/3+n pi) = sqrt(3)" "` for any integer `n`

and the only possible solutions are all of the form:

`x = pi/3 + n pi" "` for integer values of `n`.