Решебник по алгебре 11 класс Никольский Параграф 10. Равносильность уравнений на множествах Задание 19

\[\boxed{\mathbf{19.}}\]

\[\textbf{а)}\ \frac{2\sin x}{\sqrt{- x^{2} + 4x - 3}} =\]

\[= \frac{1}{\sqrt{- x^{2} + 4x - 3}}\]

\[1) - x^{2} + 4x - 3 > 0\]

\[x^{2} - 4x + 3 < 0\]

\[D_{1} = 4 - 3 = 1\]

\[x_{1} = 2 + 1 = 3;\]

\[x_{2} = 2 - 1 = 1;\]

\[(x - 1)(x - 3) < 0\]

\[- 1 < x < 3.\]

\[M = (1;3).\]

\[2)\ 2\sin x = 1\]

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

\[x_{n} = \frac{\pi}{6} + 2\pi n\ (не\ подходит);\]

\[x_{k} = \frac{5\pi}{6} + 2\pi k;\]

\[x = \frac{5\pi}{6}.\]

\[Ответ:x = \frac{5\pi}{6}.\]

\[\textbf{б)}\ \frac{2\sin x}{\sqrt{3 - 2x - x^{2}}} =\]

\[= \frac{\sqrt{2}}{\sqrt{3 - 2x - x^{2}}}\]

\[1)\ 3 - 2x - x^{2} > 0\]

\[x^{2} + 2x - 3 < 0\]

\[D_{1} = 1 + 3 = 4\]

\[x_{1} = - 1 + 2 = 1;\]

\[x_{2} = - 1 - 2 = - 3;\]

\[(x + 3)(x - 1) < 0\]

\[M = ( - 3;1).\]

\[2)\ 2\sin x = \sqrt{2}\]

\[\sin x = \frac{\sqrt{2}}{2}\]

\[x_{k} = \frac{3\pi}{4} + 2\pi k\ (не\ подходит);\]

\[x_{n} = \frac{\pi}{4} + 2\pi n;\]

\[x = \frac{\pi}{4}.\]

\[Ответ:x = \frac{\pi}{4}.\]

\[\textbf{в)}\ \frac{2\cos x}{\sqrt{\pi^{2} - 4x^{2}}} = \frac{\sqrt{2}}{\sqrt{\pi^{2} - 4x^{2}}}\]

\[1)\ \pi^{2} - 4x^{2} > 0\]

\[4x^{2} < \pi^{2}\]

\[x^{2} < \frac{\pi^{2}}{4}\]

\[x < \pm \frac{\pi}{2}.\]

\[M = \left( - \frac{\pi}{2};\frac{\pi}{2} \right).\]

\[2)\ 2\cos x = \sqrt{2}\]

\[\cos x = \frac{\sqrt{2}}{2}\]

\[x_{k} = - \frac{\pi}{4} + 2\pi k;\]

\[x_{1} = - \frac{\pi}{4}.\]

\[x_{n} = \frac{\pi}{4} + 2\pi n;\]

\[x_{2} = \frac{\pi}{4}.\]

\[Ответ:x = \pm \frac{\pi}{4}.\]

\[\textbf{г)}\ \frac{2\cos x}{\sqrt{\pi^{2} - x^{2}}} = \frac{- 1}{\sqrt{\pi^{2} - x^{2}}}\]

\[1)\ \pi^{2} - x^{2} > 0\]

\[x^{2} < \pi^{2}\]

\[x < \pm \pi.\]

\[M = ( - \pi;\ \pi).\]

\[2)\ 2\cos x = - 1\]

\[\cos x = - \frac{1}{2}\]

\[x_{n} = \frac{2\pi}{3} + 2\pi n;\ \ \]

\[x_{1} = \frac{2\pi}{3};\]

\[x_{k} = - \frac{2\pi}{3} + 2\pi n;\ \ \]

\[x_{2} = - \frac{2\pi}{3}.\]

\[Ответ:x = \pm \frac{2\pi}{3}.\]