ГДЗ по алгебре и начала математического анализа 11 класс Алимов Задание 695

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\[1)\ y = \frac{1}{2\sin^{2}x - \sin x}\]

\[2\sin^{2}x - \sin x \neq 0\]

\[\sin x \bullet \left( 2\sin x - 1 \right) \neq 0.\]

\[\sin x \neq 0\]

\[x \neq \arcsin 0 + \pi n = \pi n.\]

\[2\sin x - 1 \neq 0\]

\[2\sin x \neq 1\]

\[\sin x \neq \frac{1}{2}\]

\[x \neq ( - 1)^{n} \bullet \arcsin\frac{1}{2} + \pi n\]

\[x \neq ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n.\]

\[Ответ:\ \ x \neq \pi n;\ \ \]

\[x \neq ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n.\]

\[2)\ y = \frac{2}{\cos^{2}x - \sin^{2}x}\]

\[\cos^{2}x - \sin^{2}x \neq 0\]

\[\cos{2x} \neq 0\]

\[2x \neq \arccos 0 + \pi n\]

\[2x \neq \frac{\pi}{2} + \pi n\]

\[x \neq \frac{1}{2} \bullet \left( \frac{\pi}{2} + \pi n \right) = \frac{\pi}{4} + \frac{\text{πn}}{2}.\]

\[Ответ:\ \ x \neq \frac{\pi}{4} + \frac{\text{πn}}{2}.\]

\[3)\ y = \frac{1}{\sin x - \sin{3x}}\]

\[\sin x - \sin{3x} \neq 0\]

\[2 \bullet \sin\frac{x - 3x}{2} \bullet \cos\frac{x + 3x}{2} \neq 0\]

\[- 2 \bullet \sin x \bullet \cos{2x} \neq 0\]

\[\sin x \neq 0\]

\[x \neq \arcsin 0 + \pi n = \pi n.\]

\[\cos{2x} \neq 0\]

\[2x \neq \arccos 0 + \pi n = \frac{\pi}{2} + \pi n\]

\[x \neq \frac{1}{2} \bullet \left( \frac{\pi}{2} + \pi n \right) = \frac{\pi}{4} + \frac{\text{πn}}{2}.\]

\[Ответ:\ \ x \neq \pi n;\ \ x \neq \frac{\pi}{4} + \frac{\text{πn}}{2}.\]

\[4)\ y = \frac{1}{\cos^{3}x + \cos x}\]

\[\cos^{3}x + \cos x \neq 0\]

\[\cos x \bullet \left( \cos^{2}x + 1 \right) \neq 0\]

\[\cos x \neq 0\]

\[x \neq \arccos 0 + \pi n = \frac{\pi}{2} + \pi n.\]

\[\cos^{2}x + 1 \neq 0\]

\[\cos^{2}x \neq - 1\]

\[при\ любом\ x.\]

\[Ответ:\ \ x \neq \frac{\pi}{2} + \pi n.\]