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МатематикаГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1210
Задание 1210
\[\boxed{\mathbf{1210}\mathbf{.}}\]
\[1)\cos{2x} + 4\sin^{4}x = 8\cos^{6}x\]
\[\cos{2x} + 4 \cdot \left( \frac{1 - \cos{2x}}{2} \right)^{2} =\]
\[= 8 \cdot \left( \frac{1 + \cos{2x}}{2} \right)^{3}\]
\[t = \cos{2x}:\]
\[t + 4 \cdot \left( \frac{1 - t}{2} \right)^{2} -\]
\[- 8 \cdot \left( \frac{1 + t}{2} \right)^{3} = 0\]
\[t^{3} + 2t^{2} + 4t = 0\]
\[t\left( t^{2} + 2t + 4 \right) = 0\]
\[t = 0;\]
\[t^{2} + 2t + 4 = 0\]
\[D = 4 - 16 = - 12 < 0 \rightarrow нет\]
\[\ корней.\]
\[\cos{2x} = 0\]
\[2x = \frac{\pi}{2} + \pi k\]
\[x = \frac{\pi}{4} + \frac{\text{πk}}{2}.\]
\[Ответ:\ \frac{\pi}{4} + \frac{\text{πk}}{2}\text{.\ }\]
\[2)\cos^{4}x + \sin^{8}x = 1\]
\[\cos^{4}x + \left( 1 - \cos x \right)^{4} = 0\]
\[Пусть\ t = \cos^{2}x:\]
\[t^{2} + (1 - t)^{4} - 1 = 0\]
\[t^{4} - 4t^{3} + 7t^{2} - 4t = 0\]
\[P(0) = P(1) = 0;\]
\[t(t - 1) = t^{2} - t:\]
\[t(t - 1)\left( t^{2} - 3t + 4 \right) = 0\]
\[t = 0:\]
\[\cos^{2}x = 0\]
\[x = \frac{\pi}{2} + \pi k.\]
\[t = 1:\]
\[\cos^{2}x = 1\]
\[\cos x = \pm 1\]
\[x = \pi k.\]
\[t^{2} - 3t + 4 = 0\]
\[D = 9 - 16 = - 7 < 0 \rightarrow нет\ \]
\[корней.\]
\[Ответ:\ \frac{\pi}{2} + \pi k;\ \ \ \pi k.\]
\[3)\ 2\sin^{2}x + \frac{1}{4}\cos^{3}{2x} = 1\]
\[2\sin^{2}x - \left( \cos^{2}x + \sin^{2}x \right) +\]
\[+ \frac{1}{4}\cos^{3}{2x} = 0\]
\[- \left( \cos^{2}x - \sin^{2}x \right) +\]
\[+ \frac{1}{4}\cos^{3}{2x} = 0\]
\[- \cos{2x} + \frac{1}{4}\cos^{3}{2x} = 0\]
\[Пусть\ y = \cos{2x}:\]
\[\frac{1}{4}y^{3} - y = 0\]
\[y\left( \frac{1}{4}y^{2} - 1 \right) = 0\]
\[Первое\ уравнение:\]
\[\cos{2x} = 0\]
\[2x = \arccos 0 + \pi n = \frac{\pi}{2} + \pi n\]
\[x = \frac{1}{2} \bullet \left( \frac{\pi}{2} + \pi n \right) = \frac{\pi}{4} + \frac{\text{πn}}{2}.\]
\[Второе\ уравнение:\]
\[\frac{1}{4}\cos^{2}{2x} - 1 = 0\]
\[\frac{1}{4}\cos^{2}{2x} = 1\]
\[\cos^{2}{2x} = 4\]
\[\cos{2x} = \pm 2 - корней\ нет.\]
\[Ответ:\ \ \frac{\pi}{4} + \frac{\text{πn}}{2}.\]
\[4)\sin^{2}{2x} + \cos^{2}{3x} = 1 +\]
\[+ 4\sin x\]
\[\sin^{2}{2x} - \left( 1 - \cos^{2}{3x} \right) = 4\sin x\]
\[\sin^{2}{2x} - \sin^{2}{3x} = 4\sin x\]
\[\left( \sin{2x} + \sin{3x} \right) \bullet\]
\[\bullet \left( \sin{2x} - \sin{3x} \right) = 4\sin x\]
\[2 \bullet \sin\frac{2x + 3x}{2} \bullet \cos\frac{2x - 3x}{2} \bullet 2 \bullet\]
\[\bullet \sin\frac{2x - 3x}{2} \bullet \cos\frac{2x + 3x}{2} =\]
\[= 4\sin x\]
\[4 \bullet \sin\frac{5x}{2} \bullet \cos\left( - \frac{x}{2} \right) \bullet \sin\left( - \frac{x}{2} \right) \bullet\]
\[\bullet \cos\frac{5x}{2} = 4\sin x\]
\[- 2 \bullet \sin{5x} \bullet \cos\frac{x}{2} \bullet \sin\frac{x}{2} = 4\sin x\]
\[- \sin x \bullet \sin{5x} - 4\sin x = 0\]
\[- \sin x\left( \sin{5x} + 4 \right) = 0\]
\[Первое\ уравнение:\]
\[\sin x = 0\]
\[x = \arcsin 0 + \pi n = \pi n.\]
\[Второе\ уравнение:\]
\[\sin{5x} + 4 = 0\]
\[\sin{5x} = - 4 - корней\ нет.\]
\[Ответ:\ \ \pi n.\]
