Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1248

\[\boxed{\mathbf{1248}\mathbf{.}}\]

\[1)\sin^{2}x - \cos x \bullet \cos{3x} = \frac{1}{4}\]

\[\sin^{2}x -\]

\[- \frac{1}{2}\left( \cos(x + 3x) + \cos(x - 3x) \right) -\]

\[- \frac{1}{4} = 0\]

\[2\sin^{2}x - \left( \cos{4x} + \cos{2x} \right) -\]

\[- \frac{1}{2} = 0\]

\[2\sin^{2}x -\]

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

\[- \frac{1}{2} = 0\]

\[2\sin^{2}x -\]

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

\[- \frac{1}{2} = 0\]

\[2\sin^{2}x -\]

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

\[- \frac{1}{2} = 0\]

\[1 - \cos{2x} - \cos{2x} -\]

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

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

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

\[Пусть\ y = \cos{2x}:\]

\[4y^{2} + 4y - 3 = 0\]

\[D = 4^{2} + 4 \bullet 4 \bullet 3 = 16 +\]

\[+ 48 = 64\]

\[y_{1} = \frac{- 4 - 8}{2 \bullet 4} = - \frac{12}{8} = - \frac{3}{2}\text{\ \ }и\ \ \]

\[y_{2} = \frac{- 4 + 8}{2 \bullet 4} = \frac{4}{8} = \frac{1}{2}.\]

\[Первое\ уравнение:\]

\[\cos{2x} = - \frac{3}{2} - корней\ нет.\]

\[Второе\ уравнение:\]

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

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

\[= \pm \frac{\pi}{3} + 2\pi n\]

\[x = \frac{1}{2} \bullet \left( \pm \frac{\pi}{3} + 2\pi n \right) = \pm \frac{\pi}{6} + \pi n.\]

\[Ответ:\ \pm \frac{\pi}{6} + \pi n.\]

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

\[\sin{3x} + \sin x = 4\sin x\]

\[2 \bullet \sin\frac{3x + x}{2} \bullet \cos\frac{3x - x}{2} -\]

\[- 4\sin x = 0\]

\[2 \bullet \sin{2x} \bullet \cos x - 4\sin x = 0\]

\[4\sin x \bullet \cos x \bullet \cos x -\]

\[- 4\sin x = 0\]

\[4\sin x \bullet \left( \cos^{2}x - 1 \right) = 0\]

\[Первое\ уравнение:\]

\[\sin x = 0\]

\[x = \arcsin 0 + \pi n = \pi n.\]

\[Второе\ уравнение:\]

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

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

\[\cos x = \pm 1\]

\[x_{1} = \pi - \arccos 1 + \pi n =\]

\[= \pi + 2\pi n;\]

\[x_{2} = \arccos 1 + \pi n = 2\pi n.\]

\[Ответ:\ \ \pi n.\]

\[3)\ 3\cos{2x} - 7\sin x = 4\]

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

\[- 4 = 0\]

\[3\left( 1 - \sin^{2}x - \sin^{2}x \right) -\]

\[- 7\sin x - 4 = 0\]

\[3 - 6\sin^{2}x - 7\sin x - 4 = 0\]

\[6\sin^{2}x + 7\sin x + 1 = 0\]

\[Пусть\ y = \sin x:\]

\[6y^{2} + 7y + 1 = 0\]

\[D = 7^{2} - 6 \bullet 4 = 49 - 24 = 25\]

\[y_{1} = \frac{- 7 - 5}{2 \bullet 6} = - 1\ \ и\ \ \]

\[y_{2} = \frac{- 7 + 5}{2 \bullet 6} = \frac{2}{12} = \frac{1}{6}.\]

\[Первое\ уравнение:\]

\[\sin x = - 1\]

\[x = - \arcsin 1 + 2\pi n =\]

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

\[Второе\ уравнение:\]

\[\sin x = \frac{1}{6}\]

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

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

\[\ ( - 1)^{n} \bullet \arcsin\frac{1}{6} + \pi n.\]

\[4)\ 1 + \cos x + \cos{2x} = 0\]

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

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

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

\[\cos x \bullet \left( 2\cos x + 1 \right) = 0\]

\[Первое\ уравнение:\]

\[\cos x = 0\]

\[x = \arccos 0 + \pi n = \frac{\pi}{2} + \pi n.\]

\[Второе\ уравнение:\]

\[2\cos x + 1 = 0\]

\[2\cos x = - 1\]

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

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

\[= \pm \left( \pi - \frac{\pi}{3} \right) + 2\pi n =\]

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

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