For what numbers x, -2π≤ x ≤ 2π, does the graph of y = tan x have vertical asymptotes.?

The basic graph of the function y = tan x on -π/2 ≤ x ≤ π/2 is constrained by the asymptotes x = -π/2 (on the left) and x = π/2 (on the right. The period of the tangent function is π, so the first vertical asymptote to the left of x = -π/2 is x = -3π/2 and the first to the right of x = π/2 is 3π/2.

In summary, the vertical asymptotes of y = tan x between -2π ≤ x ≤ 2π are {-3π/2, -π/2, π/2, 3π/2}

facundo3141592 facundo3141592 · Answer 2 · 2021-08-11T15:40:30+02:00

A given function will have vertical asymptotes when its denominator becomes equal to zero.

Particularly, we know that:

tan(x) = sin(x)/cos(x)

So the x-values of the asymptote are the values of x such that:

cos(x) = 0

We know that the two zeros in the range [0, 2π] are:

cos(π/2) = 0

cos(3π/2) = 0

And the cosine function is an even function, which means that:

cos(x) = cos(-x)

So to extend to the range [-2π, 2π] we can use that property:

cos(π/2) = cos(-π/2) = 0

cos(3π/2) = cos(-3π/2) = 0

Then the four x-values of the asymptotes are:

{-3π/2, -π/2, π/2, 3π/2}

Below you can see the graph of the function tan(x)

If you want to learn more about trigonometric functions, you can read:

https://brainly.com/question/24297646