ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1599

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\[1)\cos x + \cos{2x} + \cos{3x} = 0\]

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

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

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

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

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

\[x = \frac{1}{2} \bullet \left( \frac{\pi}{2} + \pi n \right) = \frac{\pi}{4} + \frac{\text{πn}}{2}.\]

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

\[2\cos x = - 1\]

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

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

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

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

\[Ответ:\ \ \frac{\pi}{4} + \frac{\text{πn}}{2};\ \ \pm \frac{2\pi}{3} + 2\pi n.\]

\[2)\cos^{3}x - 3\cos^{2}x + \cos x + \sin{2x} =\]

\[= 2\cos\left( \frac{x}{2} + \frac{\pi}{4} \right) \bullet \sin\left( \frac{3x}{2} - \frac{\pi}{4} \right)\]

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

\[= \sin\left( \frac{3x}{2} - \frac{\pi}{4} + \frac{x}{2} + \frac{\pi}{4} \right) + \sin\left( \frac{3x}{2} - \frac{\pi}{4} - \frac{x}{2} - \frac{\pi}{4} \right)\]

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

\[= \sin{2x} + \sin\left( x - \frac{\pi}{2} \right)\]

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

\[\cos x \bullet \left( \cos^{2}x - 3\cos x + 2 \right) = 0\]

\[y = \cos x:\]

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

\[D = 3^{2} - 4 \bullet 2 = 9 - 8 = 1\]

\[y_{1} = \frac{3 - 1}{2} = 1;\]

\[y_{2} = \frac{3 + 1}{2} = 2.\]

\[1)\ \cos x = 0\]

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

\[2)\ \cos x = 1\]

\[x = \arccos 1 + 2\pi n = 2\pi n.\]

\[3)\ \cos x = 2\]

\[корней\ нет.\]

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

\[3)\sin^{2}x + \cos^{2}{3x} = 1\]

\[1 - \cos^{2}x + \cos^{2}{3x} = 1\]

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

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

\[1)\ \sin{4x} = 0\]

\[4x = \arcsin 0 + \pi n = \pi n\]

\[x = \frac{\text{πn}}{4}.\]

\[2)\ \sin{2x} = 0\]

\[2x = \arcsin 0 + \pi n = \pi n\]

\[x = \frac{\text{πn}}{2}.\]

\[Ответ:\ \ \frac{\text{πn}}{4}.\]

\[4)\ ctg\ x + \sin{2x} = ctg\ 3x\]

\[ctg\ 3x - ctg\ x = \sin{2x}\]

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

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

\[\frac{\sin(x - 3x)}{\sin x \bullet \sin x} = \sin{2x}\]

\[\frac{- \sin{2x} - \sin x \bullet \sin{2x} \bullet \sin{3x}}{\sin{3x} \bullet \sin x} = 0\]

\[- \frac{\sin{2x} \bullet \left( 1 + \sin x \bullet \sin{3x} \right)}{\sin{3x} \bullet \sin x} = 0\]

\[1)\ \sin{2x} = 0\]

\[2x = \arcsin 0 + \pi n = \pi n\]

\[x = \frac{\text{πn}}{2}.\]

\[2)\ 1 + \sin x \bullet \sin{3x} = 0\]

\[\sin x \bullet \sin{3x} = - 1\]

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

\[\cos{2x} - \cos{4x} = - 2\]

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

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

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

\[\cos^{2}{2x} + 3\left( 1 - \cos^{2}{2x} \right) + \cos{2x} = 0\]

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

\[\cos^{2}{2x} - \frac{1}{2}\cos{2x} = \frac{3}{2}\]

\[\cos^{2}{2x} - \frac{1}{2}\cos{2x} + \frac{1}{16} = \frac{25}{16}\]

\[\left( \cos{2x} - \frac{1}{4} \right)^{2} = \frac{25}{16}\]

\[\cos{2x} - \frac{1}{4} = \pm \frac{5}{4}.\]

\[1)\ \cos{2x} = - \frac{5}{4} + \frac{1}{4} = - 1\]

\[2x = \arccos( - 1) + 2\pi n\]

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

\[x = \frac{1}{2} \bullet ( - \pi + 2\pi n)\]

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

\[2)\ \cos{2x} = + \frac{5}{4} + \frac{1}{4} = \frac{3}{2}\]

\[корней\ нет.\]

\[Имеет\ смысл\ при:\]

\[\sin x \neq 0\]

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

\[\sin{3x} \neq 0\]

\[3x \neq \arcsin 0 + \pi n \neq \pi n\]

\[x \neq \frac{\text{πn}}{3}.\]

\[Ответ:\ \ \frac{\pi}{2} + \pi n;\]