ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1259

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

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

\[\frac{2\sin x\cos x - 1}{\sin x} = 0\]

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

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

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

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

\[2)\ \sqrt{6\sin x\cos{2x}} = \sqrt{- 7\sin{2x}}\]

\[6\sin x\cos{2x} = - 7\sin{2x}\]

\[2\sin x\left( 6\cos^{2}x + 7\cos x - 3 \right) =\]

\[= 0\]

\[\sin x = 0\]

\[x = \pi n.\]

\[6\cos^{2}x + 7\cos x - 3 = 0\]

\[t = \cos x:\]

\[6t^{2} + 7t - 3 = 0\]

\[D = 49 + 72 = 121\]

\[t_{1} = \frac{- 7 - 11}{12} = - \frac{18}{12} = - \frac{3}{2};\ \ \]

\[\ t_{2} = \frac{- 7 + 11}{12} = \frac{4}{12} = \frac{1}{3}.\]

\[\cos x = \frac{1}{3}\]

\[x = \pm \arccos\frac{1}{3} + 2\pi n.\]

\[Ответ:x = \pi n;\ \ \ \]

\[x = \pm \arccos\frac{1}{3} + 2\pi n.\]

\[3)\ \sqrt{5tgx + 10} = \frac{5}{2}\sin x + \frac{1}{\cos x}\]

\[5\ tg\ x + 10 = \left( \frac{5}{2}\sin x + \frac{1}{\cos x} \right)^{2}\]

\[\frac{1}{4} \cdot \frac{25\cos^{4}x + 15\cos^{2}x - 4}{\cos^{2}x} = 0\]

\[25\cos^{4}x + 15\cos^{2}x - 4 = 0\]

\[Пусть\cos^{2}x = t:\]

\[25t^{2} + 15t - 4 = 0\]

\[D = 225 + 400 = 625\]

\[t_{1} = \frac{- 15 + 25}{50} = \frac{10}{50} = \frac{1}{5};\ \ \ \]

\[t_{2} = \frac{- 15 - 25}{50} = - \frac{40}{50} = - \frac{4}{5}.\]

\[\cos^{2}x = \frac{1}{5}\]

\[\cos x = \pm \frac{1}{\sqrt{5}}\]

\[x = \pm \arccos\left( \pm \frac{1}{\sqrt{5}} \right) + 2\pi n.\]

\[Нам\ подходит:\]

\[1)\ x = \pm \arccos\frac{1}{\sqrt{5}} + 2\pi n;\]

\[2)\ x = \pi - \arccos\frac{1}{\sqrt{5}} + 2\pi n.\ \]

\[4)\ \sqrt{12 - 6\sqrt{2}\text{tgx}} =\]

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

\[12 - 6\sqrt{2}tgx =\]

\[= \left( 3\sin x - \frac{\sqrt{2}}{\cos x} \right)^{2}\]

\[\frac{9\cos^{4}x + 3\cos^{2}x - 2}{\cos^{2}x} = 0\]

\[9\cos^{4}x + 3\cos^{2}x - 2 = 0\]

\[Пусть\cos^{2}x = t:\]

\[9t^{2} + 3t - 2 = 0\]

\[D = 9 + 72 = 81\]

\[t_{1} = \frac{- 3 - 9}{18} = - \frac{12}{18} = - \frac{2}{3};\ \ \]

\[\ t_{1} = \frac{- 3 + 9}{18} = \frac{1}{18}.\]

\[\cos^{2}x = \frac{1}{3}\]

\[x = \pm \arccos\left( \pm \frac{1}{\sqrt{3}} \right) + 2\pi n.\]

\[Нам\ подходят:\]

\[x = \arccos\frac{1}{\sqrt{3}} + 2\pi n;\]

\[x = \pm \left( \pi - \arccos\frac{1}{\sqrt{3}} \right) + 2\pi n.\]