Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1372

\[\boxed{\mathbf{1372}\mathbf{.}}\]

\[1)\ tg^{3}\ x + tg^{2}\ x - 2\ tg\ x - 2 = 0\]

\[tg^{2}\ x \bullet (tg\ x + 1) - 2 \bullet (tg\ x + 1) = 0\]

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

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

\[tg^{2}\ x = 2\]

\[tg\ x = \pm \sqrt{2}\]

\[x = \pm arctg\ \sqrt{2} + \pi n.\]

\[tg\ x + 1 = 0\]

\[tg\ x = - 1\]

\[x = - arctg\ 1 + \pi n = - \frac{\pi}{4} + \pi n.\]

\[Ответ:\ \ \pm arctg\ \sqrt{2} + \pi n;\ \ \]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - \frac{\pi}{4} + \pi n.\]

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

\[1 - \cos x = \frac{\sin x}{\cos x} - \sin x\ \ | \bullet \cos x\]

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

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

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

\[1)\ \cos x - 1 = 0\]

\[\cos x = 1\]

\[x = \arccos 1 + 2\pi n = 2\pi n.\]

\[\sin x - \cos x = 0\ \ \ \ \ |\ :\cos x\]

\[tg\ x - 1 = 0\]

\[tg\ x = 1\]

\[x = arctg\ 1 + \pi n = \frac{\pi}{4} + \pi n.\]

\[Ответ:\ \ 2\pi n;\ \ \frac{\pi}{4} + \pi n.\]