Question

Найдите область определения функции
$y = \frac{1}{2sinx - 1} + \sqrt{6x - x {}^{2} }$
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Answer 1

$y = \dfrac{1}{2\sin x-1} + \sqrt{6x-x^{2}}$

Область определения:

$\left \{ {\bigg{2\sin x - 1 \neq 0 } \atop \bigg{6x-x^{2}\geq 0 \ \ \ }} \right. \ \ \ \ \ \ \ \ \ \ \ \ \left \{ {\bigg{\sin x \neq \dfrac{1}{2} \ \ \ \ \ } \atop \bigg{x(6-x) = 0}} \right. \\\\\left \{ {\bigg{x\neq (-1)^{n} \cdot \dfrac{\pi}{6} + \pi n, \ n\in Z } \atop \bigg{\left[\begin{array}{ccc}x = 0\\x = 6\end{array}\right } \bigg{\Rightarrow x \in [0; \ 6] \ \ \ \ \ \ \ } } \right.$

Пусть $n = 0$, тогда: $x\neq (-1)^{0} \cdot \dfrac{\pi}{6} + \pi \cdot 0 \neq \dfrac{\pi}{6}$

Пусть $n = 1$, тогда: $x\neq (-1)^{1} \cdot \dfrac{\pi}{6} + \pi \cdot 1 \neq \dfrac{5\pi}{6}$

Пусть $n = 2$, тогда: $x\neq (-1)^{2} \cdot \dfrac{\pi}{6} + \pi \cdot 2 \neq \dfrac{13\pi}{6}$

Ответ: $x \in [0; \ 6] \ / \ \dfrac{\pi}{6}; \ \dfrac{5\pi}{6}$