Calculate the domain of Rational function, which contains Trigonometric, Irrational and Log functions.

Question

I'm trying to find the domain of the following function:

$$f(x)={\frac{\displaystyle\ \sqrt{\frac 12-{\log_3}\biggl(\frac 12 \tan(x) + \sin(x)\biggr)} + \sqrt{\pi^2-4x^2}}{\displaystyle \arcsin\biggl(\sqrt{x^2-x} - |x|\biggr)}}$$

I have reasoned this way: since I have a Rational function, its denominator must be posed $\neq0$; the Irrational functions' argument need to be $\ge0$ and the Arcsin's argument must be $-1\le x\le 1$.

Hence I have to solve the following system: $$ \left\{ \begin{array}{c} \frac 12-{\log_3}\biggl(\frac 12 \tan(x) + \sin(x)\biggr)\ge0 \\ \frac 12 \tan(x) + \sin(x) >0\\ x\neq \frac \pi2 + k\pi\\ \pi^2-4x^2 \ge 0\\ \arcsin\biggl(\sqrt{x^2-x} - |x|\biggr) \neq 0 \\ -1\le\sqrt{x^2-x} - |x|\le1\\ x^2-x \ge0 \end{array} \right. $$

Now come the point in which I get blocked. I do not know how to solve the first inequality of the system. Specifically I proceed in this way:

I express the second member's inequality in logarithm and I obtain this situation: $${\log_3}\biggl(\frac 12 \tan(x) + \sin(x)\biggr) \le \log_3\Bigl(\sqrt3 \Bigr)\quad\mathbf{(1)}$$ Next I apply logarithm inequalities rule, and then I express $\sin(x)$ in $\tan(x)$: $$\frac 12 \tan(x)+\frac{\tan(x)}{\sqrt{1+\tan^2(x)}}\le\sqrt3$$ I try to simplify: $$\frac{\tan(x)\sqrt{1+\tan^2(x)}+2\tan(x)-2\sqrt3\biggl(\sqrt{1+\tan^2(x)}\biggr)}{2\sqrt{1+\tan^2(x)}}\le0$$

From here, I do not know how to solve this inequality.

Someone could say me if my calculus is wrong or could give me another way (or hint) to determine the domain of f(x)? Thank you.

UPDATE: The (1) became, by using logarithm inequalities rules: $$\frac 12 \tan(x) + \sin(x) \le \sqrt3$$ $$\frac 12 \frac {\sin(x)}{\cos(x)} + \sin(x) \le \sqrt3 \quad \mathbf{(2)}$$

I do this sostitution: $$t=\tan\Bigl(\frac x2\Bigr); \quad \sin(x)=\frac{2t}{1+t^2};\quad \cos(x) = \frac{1-t^2}{1+t^2}.$$

So applying this to the (2): $$\frac 12 \frac {2t}{1+t^2} \frac{1+t^2}{1-t^2} + \frac{2t}{1+t^2} \le \sqrt 3$$ $$\frac{t}{1-t^2} + \frac{2t}{1+t^2} \le \sqrt 3$$ $$\frac{t(1+t^2)+2t(1-t^2)-\sqrt{3}(1-t^4)}{(1-t^4)}\le0 \quad \mathbf{(3)}$$

I try to solve the (3), and so:

The numerator solution should be: $$t+t^3+2t-2t^3-\sqrt 3 + \sqrt 3 t^4 \ge 0$$ $$\sqrt 3 t^4 -t^3 +3t -\sqrt 3 \ge 0$$ $$\biggl(t-\frac{\sqrt 3}{3}\biggr)\biggl(t^3+\sqrt 3\biggr)\ge0$$ $$t\le-\sqrt[6]{3} \quad\lor\quad t\ge\frac{\sqrt 3}{3}$$ $$\tan\Bigl(\frac x2\Bigr)\le-\sqrt[6]{3} \quad\lor\quad \tan\Bigl(\frac x2\Bigr)\ge\frac{\sqrt 3}{3}$$ $$\frac \pi2 < \frac x2 \le \arctan(\sqrt[6]{-3}),\ in \ k\pi \quad\lor\quad \frac \pi6 \le \frac x2 < \frac \pi2,\ in \ k\pi$$ $$\pi < x < -2 \arctan(\sqrt[6]{3}),\ in \ 2k\pi \quad\lor\quad \frac \pi3 \le x < \pi,\ in\ 2k\pi$$

The denominator solution should be: $$1-t^4\ge0$$ $$-1\le t \le 1$$ $$-1\le \tan\Bigl(\frac x2\Bigr) \le 1$$ $$0\le \frac x2 \le \frac \pi4,\ in \ k\pi \quad\lor\quad \frac 34 \pi \le \frac x2 < \pi,\ in \ k\pi$$ $$0\le x\le \frac \pi2,\ in \ 2k\pi \quad\lor\quad \frac 32 \pi \le x < 2 \pi,\ in \ 2k\pi$$

But my numerator solution seems to not be right, someone has an idea or could say me where I'm wrong? Thank you.

Answer 1

May be if we use this we might get something, $$\frac{1}{2} - \log_3(\frac{1}{2}tan(x)+sin(x)) \geq 0$$ $$\Rightarrow \log_3(\frac{1}{2}tan(x)+sin(x))\leq \frac{1}{2}$$ $$\Rightarrow \frac{1}{2}tan(x)+sin(x) \leq 3^\frac{1}{2} $$ now, multiply $\sqrt{3}$ we get, $$ \frac{\sqrt{3}}{2}tan(x)+{\sqrt{3}}sin(x) \leq 3$$ $$\Rightarrow sin(\frac{\pi}{3})tan(x) + tan(\frac{\pi}{3})sin(x)\leq3 $$ I am not able to solve further.

Answer 2

1. The function domain should not be larger than the domain of the denominator. Domain of $|x|$ is $\mathbb{R}$. Domain of $\sqrt{x^2-x}$ is $\mathbb{R}$, except the interval $(0,1)$. Domain of the argument of the arcsin function is therefore $(-\infty,0] \cup [1,\infty)$. The range of this argument is $[-1,-1/2)$ or $[0,1/2)$, which belongs entirely to the domain of the arcsin function, which is $[-1,1]$. The arcsin goes to zero whenever $x$ is zero. Therefore, the domain for the whole denominator should be $(-\infty,0) \cup [1,\infty)$.
2. The square root in the numerator which contains $\pi$ limits our domain to $[-\pi/2, \pi/2]$, so our next working domain should be the intersection of the previous domain with this new domain, that is, $[-\pi/2, 0) \cup [1, \pi/2]$.
3. The argument of the $\log_3$ function is composed by two trigonometric functions: a half tangent and a sine. We know that in the last restricted working domain, the half tangent function, which is inside the log function, which only admits postive values, is only positive when $x$ is positive; the same fact occurs with the sine component, and the same with the sum. So our domain is again restricted to $[1, \pi/2)$, because of the $\log_3$ function.
4. But the whole radical that contains this log function should not be nonnegative for $[1, \pi/2)$. As stated earlier, the $\log_3$ function is all positive in such interval, but the expression is being substracted from $1/2$ in the radicand. For the radicand to be zero, $\log_3(\frac{1}{2}\tan x+\sin x)=\frac{1}{2}$, which means that $\sqrt{3}=\frac{1}{2}\tan x+\sin x$. Where is this zero? Remember some special values of trig functions and take $\pi/3$. Sine of $\pi/3$ is just $\sqrt{3}/2$. And tangent of $\pi/3$ is just $\sqrt{3}$. Then we have a zero in $\pi/3$. And this makes the radicand nonnegative when $x \leq \pi/3$.
5. That last restriction becomes the domain of the function: $[1,\pi/3]$.