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МатематикаГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1248
Задание 1248
\[\boxed{\mathbf{1248}\mathbf{.}}\]
\[1)\sin^{2}x - \cos x \bullet \cos{3x} = \frac{1}{4}\]
\[\sin^{2}x -\]
\[- \frac{1}{2}\left( \cos(x + 3x) + \cos(x - 3x) \right) -\]
\[- \frac{1}{4} = 0\]
\[2\sin^{2}x - \left( \cos{4x} + \cos{2x} \right) -\]
\[- \frac{1}{2} = 0\]
\[2\sin^{2}x -\]
\[- \left( \cos{2x} + \cos^{2}{2x} - \sin^{2}{2x} \right) -\]
\[- \frac{1}{2} = 0\]
\[2\sin^{2}x -\]
\[- \left( \cos{2x} + \cos^{2}{2x} - \left( 1 - \cos^{2}{2x} \right) \right) -\]
\[- \frac{1}{2} = 0\]
\[2\sin^{2}x -\]
\[- \left( \cos{2x} + 2\cos^{2}{2x} - 1 \right) -\]
\[- \frac{1}{2} = 0\]
\[1 - \cos{2x} - \cos{2x} -\]
\[- 2\cos^{2}{2x} + 1 - \frac{1}{2} = 0\]
\[- 2\cos^{2}{2x} - 2\cos{2x} + \frac{3}{2} = 0\]
\[4\cos^{2}{2x} + 4\cos{2x} - 3 = 0\]
\[Пусть\ y = \cos{2x}:\]
\[4y^{2} + 4y - 3 = 0\]
\[D = 4^{2} + 4 \bullet 4 \bullet 3 = 16 +\]
\[+ 48 = 64\]
\[y_{1} = \frac{- 4 - 8}{2 \bullet 4} = - \frac{12}{8} = - \frac{3}{2}\text{\ \ }и\ \ \]
\[y_{2} = \frac{- 4 + 8}{2 \bullet 4} = \frac{4}{8} = \frac{1}{2}.\]
\[Первое\ уравнение:\]
\[\cos{2x} = - \frac{3}{2} - корней\ нет.\]
\[Второе\ уравнение:\]
\[\cos{2x} = \frac{1}{2}\]
\[2x = \pm \arccos\frac{1}{2} + 2\pi n =\]
\[= \pm \frac{\pi}{3} + 2\pi n\]
\[x = \frac{1}{2} \bullet \left( \pm \frac{\pi}{3} + 2\pi n \right) = \pm \frac{\pi}{6} + \pi n.\]
\[Ответ:\ \pm \frac{\pi}{6} + \pi n.\]
\[2)\sin{3x} = 3\sin x\]
\[\sin{3x} + \sin x = 4\sin x\]
\[2 \bullet \sin\frac{3x + x}{2} \bullet \cos\frac{3x - x}{2} -\]
\[- 4\sin x = 0\]
\[2 \bullet \sin{2x} \bullet \cos x - 4\sin x = 0\]
\[4\sin x \bullet \cos x \bullet \cos x -\]
\[- 4\sin x = 0\]
\[4\sin x \bullet \left( \cos^{2}x - 1 \right) = 0\]
\[Первое\ уравнение:\]
\[\sin x = 0\]
\[x = \arcsin 0 + \pi n = \pi n.\]
\[Второе\ уравнение:\]
\[\cos^{2}x - 1 = 0\]
\[\cos^{2}x = 1\]
\[\cos x = \pm 1\]
\[x_{1} = \pi - \arccos 1 + \pi n =\]
\[= \pi + 2\pi n;\]
\[x_{2} = \arccos 1 + \pi n = 2\pi n.\]
\[Ответ:\ \ \pi n.\]
\[3)\ 3\cos{2x} - 7\sin x = 4\]
\[3\left( \cos^{2}x - \sin^{2}x \right) - 7\sin x -\]
\[- 4 = 0\]
\[3\left( 1 - \sin^{2}x - \sin^{2}x \right) -\]
\[- 7\sin x - 4 = 0\]
\[3 - 6\sin^{2}x - 7\sin x - 4 = 0\]
\[6\sin^{2}x + 7\sin x + 1 = 0\]
\[Пусть\ y = \sin x:\]
\[6y^{2} + 7y + 1 = 0\]
\[D = 7^{2} - 6 \bullet 4 = 49 - 24 = 25\]
\[y_{1} = \frac{- 7 - 5}{2 \bullet 6} = - 1\ \ и\ \ \]
\[y_{2} = \frac{- 7 + 5}{2 \bullet 6} = \frac{2}{12} = \frac{1}{6}.\]
\[Первое\ уравнение:\]
\[\sin x = - 1\]
\[x = - \arcsin 1 + 2\pi n =\]
\[= - \frac{\pi}{2} + 2\pi n.\]
\[Второе\ уравнение:\]
\[\sin x = \frac{1}{6}\]
\[x = ( - 1)^{n} \bullet \arcsin\frac{1}{6} + \pi n.\]
\[Ответ:\ - \frac{\pi}{2} + 2\pi n;\ \]
\[\ ( - 1)^{n} \bullet \arcsin\frac{1}{6} + \pi n.\]
\[4)\ 1 + \cos x + \cos{2x} = 0\]
\[\cos^{2}x + \sin^{2}x + \cos x +\]
\[+ \cos^{2}x - \sin^{2}x = 0\]
\[2\cos^{2}x + \cos x = 0\]
\[\cos x \bullet \left( 2\cos x + 1 \right) = 0\]
\[Первое\ уравнение:\]
\[\cos x = 0\]
\[x = \arccos 0 + \pi n = \frac{\pi}{2} + \pi n.\]
\[Второе\ уравнение:\]
\[2\cos x + 1 = 0\]
\[2\cos x = - 1\]
\[\cos x = - \frac{1}{2}\]
\[x = \pm \left( \pi - \arccos\frac{1}{2} \right) + 2\pi n =\]
\[= \pm \left( \pi - \frac{\pi}{3} \right) + 2\pi n =\]
\[= \pm \frac{2\pi}{3} + 2\pi n.\]
\[Ответ:\ \ \frac{\pi}{2} + \pi n;\ \ \pm \frac{2\pi}{3} + 2\pi n.\]
