Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 880

\[1)\sin x + \cos x = \sqrt{1 + tg\ x}\]

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

\[= 1 + tg\ x\]

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

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

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

\[1)\ \sin x = 0\]

\[x = \pi n.\]

\[2)\ 2\cos^{2}x - 1 = 0\]

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

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

\[x = \frac{\pi}{4} + \frac{\text{πn}}{2}.\]

\[Область\ определения:\]

\[\sin x + \cos x \geq 0\]

\[tg\ x + 1 \geq 0\]

\[tg\ x \geq - 1\]

\[- \frac{\pi}{4} + \pi n \leq x < \frac{\pi}{2} + \pi n.\]

\[Ответ:\ \ 2\pi n;\ \frac{3\pi}{4} + \pi n;\ \frac{\pi}{4} + 2\pi n.\]

\[2)\ \sqrt{5\sin{2x} - 2} = \sin x - \cos x\]

\[5\sin{2x} - 2 =\]

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

\[5\sin{2x} - 2 = 1 - \sin{2x}\]

\[6\sin{2x} = 3\]

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

\[2x = ( - 1)^{n} \bullet \arcsin\frac{1}{2} + \pi n\]

\[2x = ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n\]

\[x = \frac{1}{2}\left( ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n \right)\]

\[x = ( - 1)^{n} \bullet \frac{\pi}{12} + \frac{\text{πn}}{2}.\]

\[Область\ определения:\]

\[\sin x - \cos x \geq 0\]

\[5\sin{2x} - 2 \geq 0\]

\[\sin{2x} \geq \frac{2}{5}.\]

\[Ответ:\ \ \frac{5\pi}{12} + 2\pi n;\ \frac{13\pi}{12} + 2\pi n.\]