<s>[INST] Calculate the domain of Rational function, which contains Trigonometric, Irrational and Log functions. 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.
 [/INST] 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.
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