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find the vertical asymptotes of function f(`theta)=6tan(1/3 theta )`

The equation of the vertical asymptote of the given function is x = (3pi/2) +3pi*n, where n is any integer.

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The given trigonometric function is:

`f(theta) = 6tan(1/3 theta)`

Take note that a tangent function in its basic form `y=tan(x)` has a vertical asymptote at `x=pi/2` and `x = (3pi)/2` on the interval `[0,2pi]` .

So to solve for the vertical asymptote of the given tangent function set the expression inside the parenthesis equal to `pi/2` and `(3pi)/2` . And solve for theta.

`1/3 theta = pi/2`

`theta = (3pi)/2`

`1/3 theta = (3pi)/2`

`theta = (9pi)/2`

Notice that on the interval `[0,2pi],` the function has only one vertical asymptote which is `theta = (3pi)/2` .

To get the vertical asymptote of the function on the interval `(-oo, oo)` , determine the period of the function. For tangent function, the formula in finding the period is:

`period = pi/B`

where B is the coefficient of theta.

Plugging in the value of B, the period of the given function is:

`period = pi/(1/3)=3pi`

Therefore, the vertical asymptote of the given tangent function is `theta = (3pi)/2 + 3pin` , where n is any integer.

(See attached graph.)

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