ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1197

\[\boxed{\mathbf{1197}\mathbf{.}}\]

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

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

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

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

\[x - \frac{\pi}{4} = \pm \arccos\frac{\sqrt{2}}{2} + 2\pi n =\]

\[= \pm \frac{\pi}{4} + 2\pi n\]

\[x_{1} = - \frac{\pi}{4} + 2\pi n + \frac{\pi}{4} = 2\pi n;\]

\[x_{2} = + \frac{\pi}{4} + 2\pi n + \frac{\pi}{4} = \frac{\pi}{2} + 2\pi n.\]

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

\[2)\ \sin{3x} + \cos{3x} = \sqrt{2}\text{\ \ \ }|\ :\sqrt{2}\]

\[\frac{\sqrt{2}}{2} \bullet \sin{3x} + \cos{3x} \bullet \frac{\sqrt{2}}{2} = 1\]

\[\cos\frac{\pi}{4} \bullet \sin{3x} + \cos{3x} \bullet \sin\frac{\pi}{4} = 1\]

\[\sin\left( \frac{\pi}{4} + 3x \right) = 1\]

\[3x + \frac{\pi}{4} = \arcsin 1 + 2\pi n =\]

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

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

\[x = \frac{1}{3} \bullet \left( \frac{\pi}{4} + 2\pi n \right) = \frac{\pi}{12} + \frac{2\pi n}{3}.\]

\[Ответ:\ \ \frac{\pi}{12} + \frac{2\pi n}{3}.\]

\[3)\ 10\sin^{2}x + 5\sin x\cos x +\]

\[+ \cos^{2}x = 3\]

\[10\sin^{2}x + 5\sin x\cos x +\]

\[+ \cos^{2}x - 3 \cdot\]

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

\[7\sin^{2}x + 5\sin x\cos x -\]

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

\[7tg^{2}x + 5tg\ x - 2 = 0\]

\[Пусть\ tg\ x = y:\]

\[7y^{2} + 5y - 2 = 0\]

\[D = 25 + 56 = 81\]

\[y_{1} = \frac{- 5 + 9}{14} = \frac{4}{14} = \frac{2}{7};\ \ \]

\[\text{\ \ }y_{2} = \frac{- 5 - 9}{14} = - 1.\]

\[1)\ tgx = \frac{2}{7}\]

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

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

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

\[Ответ:\ arctg\frac{2}{7} + \pi k;\ - \frac{\pi}{4} + \pi k.\]

\[4)\ 6\sin^{2}{2x} + 4\cos^{2}{2x} -\]

\[- 8\sin{2x}\cos{2x} = 1\]

\[6\sin^{2}{2x} + 4\cos^{2}{2x} -\]

\[- 8\sin{2x}\cos{2x} -\]

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

\[5\sin^{2}{2x} + 3\cos^{2}{2x} -\]

\[- 8\sin{2x}\cos{2x} = 0\ \ |\ :\cos^{2}{2x}\]

\[5tg^{2}2x - 8tg\ 2x + 3 = 0\]

\[Пусть\ tg\ 2x = y:\]

\[5y^{2} - 8y + 3 = 0\]

\[D_{1} = 16 - 15 = 1\]

\[y_{1} = \frac{4 + 1}{5} = 1;\ \ \ \]

\[y_{2} = \frac{4 - 1}{5} = \frac{3}{5}.\]

\[1)\ tg\ 2x = 1\]

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

\[x = \frac{\pi}{8} + \frac{\text{πk}}{2}.\]

\[2)\ tg\ 2x = \frac{3}{5}\]

\[2x = arctg\frac{3}{5} + \pi k\]

\[x = \frac{1}{2}\text{arctg}\frac{3}{5} + \frac{\text{πk}}{2}.\]

\[Ответ:\ \frac{\pi}{8} + \frac{\text{πk}}{2};\ \ \]

\[\ \frac{1}{2}\text{arctg}\frac{3}{5} + \frac{\text{πk}}{2}.\]