Find domain of tanx/tan3x
the function is undefined where the denominator is zero. That is,
tan(3x) = 0
this occurs when 3x = nπ
or, x = n * π/3
where x=nπ the function is 0/0, so there are holes there.
the function is also undefined when the numerator or denominator is undefined. That is, when
3x or x is an odd multiple of π/2
If you look at the graph, all you see are the asymptotes, but the holes are also excluded from the domain.
To find the domain of the function f(x) = tan(x)/tan(3x), we need to consider the restrictions on the values of x where the function is defined.
The tangent function, tan(x), is defined for all real numbers except for the values where the denominator is zero, tan(3x) ≠ 0. The tangent function is undefined at these points.
To find where tan(3x) ≠ 0, we can determine the values of x that would make the numerator and denominator of tan(3x) equal to zero separately.
For the numerator, tan(x), it is undefined when x = (n + 0.5)π, where n is an integer.
For the denominator, tan(3x) = 0 when 3x = nπ, where n is an integer. Solving for x, we have x = (nπ)/3.
By comparing the two conditions, we can see that the values (n + 0.5)π should not equal (nπ)/3 for the function to be defined.
Simplifying this inequality, we get:
(n + 0.5)π ≠ (nπ)/3
Multiplying both sides by 3, we have:
3(n + 0.5)π ≠ nπ
Expanding and simplifying, we get:
3nπ + 1.5π ≠ nπ
Rearranging terms, we have:
2nπ ≠ 1.5π
Dividing both sides by π, we have:
2n ≠ 1.5
Simplifying further, we find:
n ≠ 0.75
So, the domain of f(x) = tan(x)/tan(3x) is all real numbers except for x = (n + 0.5)π and x = (nπ)/3, where n is an integer, and x ≠ (0.75/2)π.
tanx/tan3x
=I1−3tan^2x)/(3−tan^2x)_
=3(3−tan^2x)−8)/(3−tan^2x)
=(1-3tan^2x)/(3−tan^2x)
Now let z = tanx
f(z)=(1-3z^2/(3-z^2)
so it looks like z cannot be sqrt(3), or
tan x cannot be sqrt3, or x cannot be....