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МатематикаРешебник по алгебре и начала математического анализа 11 класс Колягин Задание 874
Задание 874
\[1)\sin x + \sin{2x} + \sin{3x} + \sin{4x} = 0\]
\[\sin\frac{5x}{2} \bullet \cos\frac{3x}{2} + \sin\frac{5x}{2} + \cos\frac{x}{2} = 0\]
\[\sin\frac{5x}{2} \bullet \left( \cos\frac{3x}{2} + \cos\frac{x}{2} \right) = 0\]
\[\sin\frac{5x}{2} \bullet 2 \bullet \cos\frac{\frac{3x}{2} + \frac{x}{2}}{2} \bullet \cos\frac{\frac{3x}{2} - \frac{x}{2}}{2} = 0\]
\[\sin\frac{5x}{2} \bullet \cos x \bullet \cos\frac{x}{2} = 0\]
\[1)\ \sin\frac{5x}{2} = 0\]
\[\frac{5x}{2} = \pi n\]
\[x = \frac{2\pi n}{5}.\]
\[2)\ \cos x = 0\]
\[x = \frac{\pi}{2} + \pi n.\]
\[3)\ \cos\frac{x}{2} = 0\]
\[\frac{x}{2} = \frac{\pi}{2} + \pi n\]
\[x = \pi + 2\pi n\]
\[Ответ:\ \ \frac{2\pi n}{5};\ \frac{\pi}{2} + \pi n;\ \pi + 2\pi n.\]
\[2)\cos x + \cos{2x} + \cos{3x} + \cos{4x} = 0\]
\[\cos\frac{5x}{2} \bullet \cos\frac{3x}{2} + \cos\frac{5x}{2} \bullet \cos\frac{x}{2} = 0\]
\[\cos\frac{5x}{2} \bullet \left( \cos\frac{3x}{2} + \cos\frac{x}{2} \right) = 0\]
\[\cos\frac{5x}{2} \bullet 2 \bullet \cos\frac{\frac{3x}{2} + \frac{x}{2}}{2} \bullet \cos\frac{\frac{3x}{2} - \frac{x}{2}}{2} = 0\]
\[\cos\frac{5x}{2} \bullet \cos x \bullet \cos\frac{x}{2} = 0\]
\[1)\ \cos\frac{5x}{2} = 0\]
\[\frac{5x}{2} = \frac{\pi}{2} + \pi n\]
\[x = \frac{\pi}{5} + \frac{2\pi n}{5}.\]
\[2)\ \cos x = 0\]
\[x = \frac{\pi}{2} + \pi n.\]
\[3)\ \cos\frac{x}{2} = 0\]
\[\frac{x}{2} = \frac{\pi}{2} + \pi n\]
\[x = \pi + 2\pi n.\]
\[Ответ:\ \ \frac{\pi}{5} + \frac{2\pi n}{5};\ \frac{\pi}{2} + \pi n.\]
\[3)\cos x \bullet \cos{3x} = - 0,5\]
\[\cos x \bullet \left( 4\cos^{3}x - 3\cos x \right) = - \frac{1}{2}\]
\[8\cos^{4}x - 6\cos^{2}x + 1 = 0\]
\[y = \cos^{2}x:\]
\[8y^{2} - 6y + 1 = 0\]
\[D = 36 - 32 = 4\]
\[y_{1} = \frac{6 - 2}{2 \bullet 8} = \frac{1}{4};\]
\[y_{2} = \frac{6 + 2}{2 \bullet 8} = \frac{1}{2}.\]
\[1)\ \cos^{2}x = \frac{1}{4}\]
\[1 - \sin^{2}x = \frac{1}{4}\]
\[\sin^{2}x = \frac{3}{4}\]
\[\sin x = \pm \frac{\sqrt{3}}{2}\]
\[x = \pm \arcsin\frac{\sqrt{3}}{2} + \pi n = \pm \frac{\pi}{3} + \pi n.\]
\[2)\ \cos^{2}x = \frac{1}{2}\]
\[1 - \sin^{2}x = \frac{1}{2}\]
\[\sin^{2}x = \frac{1}{2}\]
\[\sin x = \pm \frac{\sqrt{2}}{2}\]
\[x = \pm \arcsin\frac{\sqrt{2}}{2} + \pi n = \pm \frac{\pi}{4} + \pi n.\]
\[Ответ:\ \pm \frac{\pi}{3} + \pi n;\ \pm \frac{\pi}{4} + \pi n.\]
