Решебник по алгебре 11 класс Никольский Параграф 10. Равносильность уравнений на множествах Задание 45

\[\boxed{\mathbf{45.}}\]

\[\textbf{а)}\ \frac{1}{\sqrt{1 - x^{2}}} = \left( 1 - x^{2} \right)^{- \frac{\text{si}n^{2}x}{\sin x + 1}}\]

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

\[x^{2} - 1 < 0\]

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

\[- 1 < x < 1.\]

\[\sin x + 1 \neq 0\]

\[\sin x \neq - 1\]

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

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

\[1)\lg\left( 1 - x^{2} \right) = 0\]

\[1 - x^{2} = 1\]

\[x^{2} = 0\]

\[x = 0.\]

\[2)\ \frac{\text{si}n^{2}x}{\sin x + 1} = \frac{1}{2}\]

\[\sin x = t:\]

\[\frac{t^{2}}{t + 1} = \frac{1}{2}\]

\[2t^{2} = t + 1\]

\[2t^{2} - t - 1 = 0\]

\[D = 1 + 8 = 9\]

\[t_{1} = \frac{1 + 3}{4} = 1;\]

\[t_{2} = \frac{1 - 3}{4} = - \frac{1}{2}.\]

\[\sin x = 1\]

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

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

\[x = - \frac{\pi}{6} + 2\pi n.\]

\[Ответ:x = - \frac{\pi}{6};\ \ x = 0.\]

\[\textbf{б)}\ \frac{1}{\sqrt{4 - x^{2}}} = \left( 4 - x^{2} \right)^{- \frac{\text{co}s^{2}x}{\cos x + 1}}\]

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

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

\[(x + 2)(x - 2) < 0\]

\[- 2 < x < 2.\]

\[\cos x + 1 \neq 0\]

\[\cos x \neq - 1\]

\[x \neq \pi + 2\pi n.\]

\[M = ( - 2;2).\]

\[1)\lg\left( 4 - x^{2} \right) = 0\]

\[4 - x^{2} = 1\]

\[x^{2} = 3\]

\[x = \pm \sqrt{3}.\]

\[2)\ \frac{\text{co}s^{2}x}{\cos x + 1} = \frac{1}{2}\]

\[\cos x = t:\]

\[\frac{t^{2}}{t + 1} = \frac{1}{2}\]

\[2t^{2} = t + 1\]

\[2t^{2} - t - 1 = 0\]

\[D = 1 + 8 = 9\]

\[t_{1} = \frac{1 + 3}{4} = 1;\]

\[t_{2} = \frac{1 - 3}{4} = - \frac{1}{2}.\]

\[\cos x = 1\]

\[x = 2\pi k.\]

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

\[x = \pm \frac{2\pi}{3} + 2\pi n.\]

\[Ответ:x = 0;x = \pm \sqrt{3}.\]

\[\textbf{в)}\ \frac{1}{\sqrt{2 - x^{2}}} = \left( 2 - x^{2} \right)^{- \frac{\text{si}n^{2}x}{\sin x + 1}}\]

\[2 - x^{2} > 0\]

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

\[\left( x + \sqrt{2} \right)\left( x - \sqrt{2} \right) < 0\]

\[- \sqrt{2} < x < \sqrt{2}.\]

\[\sin x + 1 \neq 0\]

\[\sin x \neq - 1\]

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

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

\[1)\lg\left( 2 - x^{2} \right) = 0\]

\[2 - x^{2} = 1\]

\[x^{2} = 1\]

\[x = \pm 1.\]

\[2)\ \frac{\text{si}n^{2}x}{\sin x + 1} = \frac{1}{2}\]

\[\sin x = t:\]

\[\frac{t^{2}}{t + 1} = \frac{1}{2}\]

\[2t^{2} = t + 1\]

\[2t^{2} - t - 1 = 0\]

\[D = 1 + 8 = 9\]

\[t_{1} = \frac{1 + 3}{4} = 1;\]

\[t_{2} = \frac{1 - 3}{4} = - \frac{1}{2}.\]

\[\sin x = 1\]

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

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

\[x = - \frac{\pi}{6} + 2\pi n.\]

\[Ответ:x = - \frac{\pi}{6};\ \ x = \pm 1.\]

\[\textbf{г)}\ \frac{1}{\sqrt{2 - x^{2}}} = \left( 2 - x^{2} \right)^{- \frac{\text{co}s^{2}x}{\cos x + 1}}\]

\[2 - x^{2} > 0\]

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

\[\left( x + \sqrt{2} \right)\left( x - \sqrt{2} \right) < 0\]

\[- \sqrt{2} < x < \sqrt{2}.\]

\[\cos x + 1 \neq 0\]

\[\cos x \neq - 1\]

\[x \neq \pi + 2\pi n.\]

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

\[1)\lg\left( 2 - x^{2} \right) = 0\]

\[2 - x^{2} = 1\]

\[x^{2} = 1\]

\[x = \pm 1.\]

\[2)\ \frac{\text{co}s^{2}x}{\cos x + 1} = \frac{1}{2}\]

\[\cos x = t:\]

\[\frac{t^{2}}{t + 1} = \frac{1}{2}\]

\[2t^{2} = t + 1\]

\[2t^{2} - t - 1 = 0\]

\[D = 1 + 8 = 9\]

\[t_{1} = \frac{1 + 3}{4} = 1;\]

\[t_{2} = \frac{1 - 3}{4} = - \frac{1}{2}.\]

\[\cos x = 1\]

\[x = 2\pi k.\]

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

\[x = \pm \frac{2\pi}{3} + 2\pi n.\]

\(Ответ:x = 0;x = \pm 1.\)