Решебник по алгебре 11 класс Никольский Параграф 8. Уравнения-следствия Задание 30

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

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

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

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

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

\[1)\cos x = 0\]

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

\[2)\cos x = 1\]

\[x = 2\pi k.\]

\[Проверка.\]

\[x = \frac{\pi}{2} + \pi k:\]

\[\text{tg\ }\left( \frac{\pi}{2} + \pi k \right) - не\ существует.\]

\[x = 2\pi k:\]

\[\frac{1 - tg^{2}(2\pi k)}{1 + tg^{2}(2\pi k)} =\]

\[= \cos(2\pi k) - \sin^{2}(2\pi k)\]

\[\frac{1 - 0}{1 + 0} = 1 - 0\]

\[1 = 1.\]

\[Ответ:x = 2\pi k.\]

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

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

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

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

\[\sin x\left( \sin x + 1 \right) = 0\]

\[1)\sin x = 0\]

\[x = \pi k.\]

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

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

\[Проверка.\]

\[x = \pi k:\]

\[\frac{1 - tg^{2}(\pi k)}{1 + tg^{2}(\pi k)} = \cos^{2}\text{πk} + \sin\text{πk}\]

\[\frac{1 - 0}{1 + 0} = 1 + 0\]

\[1 = 1.\]

\[x = - \frac{\pi}{2} + 2\pi k:\]

\[\text{tg}\left( - \frac{\pi}{2} + 2\pi k \right) - не\ существует.\]

\[Ответ:x = \text{πk.}\]

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

\[\sin{2x} = 2\cos^{2}x - \sqrt{2}\cos x\]

\[2\sin x\cos x = 2\cos^{2}x - \sqrt{2}\cos x\]

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

\[\cos x\left( 2\sin x - 2\cos x + \sqrt{2} \right) = 0\]

\[1)\cos x = 0\]

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

\[2)\ 2\sin x - 2\cos x + \sqrt{2} = 0\]

\[2\sin x - 2\cos x = - \sqrt{2}\ \ \ |\ :2\sqrt{2}\]

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

\[\sin\frac{\pi}{4}\sin x - \cos\frac{\pi}{4}\cos x = - \frac{1}{2}\]

\[\cos\left( x + \frac{\pi}{4} \right) = \frac{1}{2}\]

\[x + \frac{\pi}{4} = \pm \frac{\pi}{3} + 2\pi k\]

\[x_{1} = \frac{\pi}{12} + 2\pi k;\]

\[x_{2} = - \frac{7\pi}{12} + 2\pi k.\]

\[tg^{2}x \neq 0\]

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

\[Ответ:x = \frac{\pi}{12} + 2\pi k;\]

\[x = - \frac{7\pi}{12} + 2\pi k.\]

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

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

\[2\sin x = \sqrt{2}\cos x - 2\cos^{2}x\]

\[2\sin x\cos x = \sqrt{2}\cos x - 2\cos^{2}x\]

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

\[\cos x\left( 2\sin x - \sqrt{2} + 2\cos x \right) = 0\]

\[1)\cos x = 0\]

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

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

\[\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x = \frac{1}{2}\]

\[\sin\frac{\pi}{4}\sin x + \cos\frac{\pi}{4}\cos x = \frac{1}{2}\]

\[\cos\left( x - \frac{\pi}{4} \right) = \frac{1}{2}\]

\[x - \frac{\pi}{4} = \pm \frac{\pi}{3} + 2\pi k\]

\[x_{1} = - \frac{\pi}{12} + 2\pi k;\]

\[x_{2} = \frac{7\pi}{12} + 2\pi k.\]

\[tg^{2}x \neq 0\]

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

\[Ответ:x = - \frac{\pi}{12} + 2\pi k;\]

\[x = \frac{7\pi}{12} + 2\pi k.\]