Решебник по алгебре 11 класс Никольский Параграф 10. Равносильность уравнений на множествах Задание 30

\[\boxed{\mathbf{30.}}\]

\[\textbf{а)}\ \frac{1 - tg^{2}x}{1 + tg^{2}x} = \sin x\]

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

\[\cos{2x} = \sin x\]

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

\[2sin^{2}x + \sin x - 1 = 0\]

\[t = \sin x:\]

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

\[D = 1 + 8 = 9\]

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

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

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

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

\[x_{m} = \frac{5\pi}{6} + 2\pi m.\]

\[\sin x = - 1\]

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

\[Ответ:\ \frac{\pi}{6} + 2\pi k;\ \ \frac{5\pi}{6} + 2\pi m.\]

\[\textbf{б)}\ \frac{1 - tg^{2}x}{1 + tg^{2}x} = \cos x - 1\]

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

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

\[2cos^{2}x - \cos x = 0\]

\[\cos x\left( 2\cos x - 1 \right) = 0\]

\[\cos x = 0\]

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

\[2\cos x = 1\]

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

\[x_{k} = \pm \frac{\pi}{3} + 2\pi k.\]

\[Ответ:x = \pm \frac{\pi}{3} + 2\pi k.\]

\[\textbf{в)}\ \frac{2tgx}{1 + tg^{2}x} = \cos x\]

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

\[\sin{2x} = \cos x\]

\[2\sin x\cos x = \cos x\]

\[\cos x\left( 2\sin x - 1 \right) = 0\]

\[\cos x = 0\]

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

\[2\sin x = 1\]

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

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

\[Ответ:x = ( - 1)^{n}\frac{\pi}{6} + \pi n.\]

\[\textbf{г)}\ \frac{2tgx}{1 + tg^{2}x} = \sin x\]

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

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

\[2\sin x\cos x = \sin x\]

\[\sin x\left( 2\cos x - 1 \right) = 0\]

\[\sin x = 0\]

\[x = \pi n.\]

\[2\cos x = 1\]

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

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

\[Ответ:x = \pi n;x = \pm \frac{\pi}{3} + 2\pi k.\]