Алгебра
Английский
Биология
География
Геометрия
История
Русский
Обществознание
Физика
Химия
Информатика
Литература
МатематикаРешебник по алгебре и начала математического анализа 10 класс Колягин Задание 1206
Задание 1206
\[\boxed{\mathbf{1206}\mathbf{.}}\]
\[1)\ 2\sin^{2}x = 1 + \frac{1}{3}\sin{4x}\]
\[1 - \cos{2x} = 1 +\]
\[+ \frac{2}{3}\sin{2x} \bullet \cos{2x}\]
\[\frac{2}{3}\sin{2x} \bullet \cos{2x} + \cos{2x} = 0\]
\[\cos{2x} \bullet \left( \frac{2}{3}\sin{2x} + 1 \right) = 0\]
\[Первое\ уравнение:\]
\[\cos{2x} = 0\]
\[2x = \arccos 0 + \pi n = \frac{\pi}{2} + \pi n\]
\[x = \frac{1}{2} \bullet \left( \frac{\pi}{2} + \pi n \right) = \frac{\pi}{4} + \frac{\text{πn}}{2}.\]
\[Второе\ уравнение:\]
\[\frac{2}{3}\sin{2x} + 1 = 0\]
\[\frac{2}{3}\sin{2x} = - 1\]
\[\sin{2x} = - \frac{3}{2} - корней\ нет.\]
\[Ответ:\ \ \frac{\pi}{4} + \frac{\text{πn}}{2}.\]
\[2)\ 2\cos^{2}{2x} - 1 = \sin{4x}\]
\[1 + \cos{4x} - 1 - \sin{4x} = 0\]
\[\cos{4x} - \sin{4x} = 0\ \ \ \ \ |\ :\cos{4x}\]
\[1 - tg\ 4x = 0\]
\[tg\ 4x = 1\]
\[4x = arctg\ 1 + \pi n = \frac{\pi}{4} + \pi n\]
\[x = \frac{1}{4} \bullet \left( \frac{\pi}{4} + \pi n \right) = \frac{\pi}{16} + \frac{\text{πn}}{4}\]
\[Ответ:\ \ \frac{\pi}{16} + \frac{\text{πn}}{4}.\]
\[3)\ 2\cos^{2}{2x} + 3\cos^{2}x = 2\]
\[2\cos^{2}{2x} + \frac{3}{2}\left( 1 + \cos{2x} \right) = 2\]
\[2\cos^{2}{2x} + 1,5 + 1,5\cos{2x} -\]
\[- 2 = 0\]
\[2\cos^{2}{2x} + 1,5\cos{2x} - 0,5 = 0\]
\[Пусть\ y = \cos{2x}:\]
\[2y^{2} + 1,5y - 0,5 = 0\]
\[4y^{2} + 3y - 1 = 0\]
\[D = 3^{2} + 4 \bullet 4 = 9 + 16 = 25\]
\[y_{1} = \frac{- 3 - 5}{2 \bullet 4} = - 1\ \ и\ \]
\[\ y_{2} = \frac{- 3 + 5}{2 \bullet 4} = \frac{1}{4}.\]
\[Первое\ уравнение:\]
\[\cos{2x} = - 1\]
\[2x = \pi - \arccos 1 + 2\pi n =\]
\[= \pi + 2\pi n\]
\[x = \frac{1}{2} \bullet (\pi + \pi n) = \frac{\pi}{2} + \pi n.\]
\[Второе\ уравнение:\]
\[\cos{2x} = \frac{1}{4}\]
\[2x = \pm \arccos\frac{1}{4} + 2\pi n\]
\[x = \frac{1}{2} \bullet \left( \pm \arccos\frac{1}{4} + 2\pi n \right) =\]
\[= \pm \frac{1}{2}\arccos\frac{1}{4} + \pi n.\]
\[Ответ:\ \ \frac{\pi}{2} + \pi n;\ \]
\[\ \pm \frac{1}{2}\arccos\frac{1}{4} + \pi n.\]
\[4)\ \left( \sin x + \cos x \right)^{2} = 1 + \cos x\]
\[\cos^{2}x + \sin^{2}x + 2\sin x \bullet \cos x =\]
\[= 1 + \cos x\]
\[1 + 2\sin x \bullet \cos x - 1 -\]
\[- \cos x = 0\]
\[2\sin x \bullet \cos x - \cos x = 0\]
\[\cos x \bullet \left( 2\sin x - 1 \right) = 0\]
\[Первое\ уравнение:\]
\[\cos x = 0\]
\[x = \arccos 0 + \pi n = \frac{\pi}{2} + \pi n.\]
\[Второе\ уравнение:\]
\[2\sin x - 1 = 0\]
\[2\sin x = 1\]
\[\sin x = \frac{1}{2}\]
\[x = ( - 1)^{n} \bullet \arcsin\frac{1}{2} + \pi n =\]
\[= ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n.\]
\[Ответ:\ \ \frac{\pi}{2} + \pi n;\ \ ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n.\]
