# Domain of One Variable Function – A function with sin inside ln – Exercise 2451

Exercise

Determine the domain of the function:

$$f(x)=\ln (\sin \frac{\pi}{x})$$

$$\frac{1}{1+2 k}< x<\frac{1}{ 2k}, k>0$$

$$\frac{1}{1+2 k}> x>\frac{1}{ 2k}, k<0$$

$$x>1, k=0$$

Solution

Let’s find the domain of the function:

$$f(x)=\ln (\sin \frac{\pi}{x})$$

Because there is a denominator, the denominator must be different from zero:

$$x\neq 0$$

Also, there is a ln function, so we need the expressions inside the ln to be positive:

$$\sin \frac{\pi}{x}> 0$$

Let’s look at inequality. Sin function is positive in the first half of its cycle. Therefore, in the first cycle we receive:

$$0<\frac{\pi}{x}<\pi$$

But it has endless cycles. We express this with parameter k (integer). Adding the cycle of sin function in both sides results in

$$2\pi k < \frac{\pi}{x}< \pi+2\pi k$$

$$2\pi k< \frac{\pi}{x}< \pi (1+2k)$$

$$2 k< \frac{1}{x}< 1+2k$$

Now one see that to isolate x one have to break up into cases:

where k = 0 we get

$$0< \frac{1}{x}< 1$$

From this inequality we get that for x> 0 (positive) we get

$$x>0\Longrightarrow 0<11$$

But for x <0 (negative) we get

$$x<0\Longrightarrow 0>1>x$$

Meaning, there is no solution.

Remember that x is different from zero, otherwise we get a denominator function equal to zero, which is not defined.

Let’s move to find thedomain for k that is different from zero. Recall the inequalities we received:

$$2 k< \frac{1}{x}< 1+2k$$

Again, to divide by x or k one have to determine whether they are positive or negative. Therefore, when k> 0 (positive) then for x <0 (negative) we get

$$\frac{1}{2 k}< x<\frac{1}{ 1+2k}$$

That is, there is no solution. But for x> 0 (positive) we get

$$\frac{1}{1+2 k}< x<\frac{1}{ 2k}$$

And when k <0 then for x> 0 we get

$$\frac{1}{2 k}> x>\frac{1}{1+ 2k}$$

That is, there is no solution. But for x <0 we get

$$\frac{1}{1+2 k}>x>\frac{1}{ 2k}$$

And the final answer is the union of the 3 results we got: for k = 0, k> 0 and k <0. Therefore, we get

$$\frac{1}{1+2 k}< x<\frac{1}{ 2k}, k>0$$

and

$$\frac{1}{1+2 k}> x>\frac{1}{ 2k}, k<0$$

and

$$x>1, k=0$$

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