Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 627

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\[1)\cos{3x} - \cos{5x} = \sin{4x}\]

\[- 2 \bullet \sin\frac{3x + 5x}{2} \bullet \sin\frac{3x - 5x}{2} = \sin{4x}\]

\[- 2 \bullet \sin\frac{8x}{2} \bullet \sin\left( - \frac{2x}{2} \right) - \sin{4x} = 0\]

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

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

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

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

\[x = \frac{1}{4} \bullet \pi n = \frac{\text{πn}}{4}.\]

\[2)\ 2\sin x - 1 = 0\]

\[2\sin x = 1\]

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

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

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

\[Ответ:\ \ \frac{\text{πn}}{4};\ \ ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n.\]

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

\[2 \bullet \sin\frac{7x - x}{2} \bullet \cos\frac{7x + x}{2} = \cos{4x}\]

\[2 \bullet \sin\frac{6x}{2} \bullet \cos\frac{8x}{2} - \cos{4x} = 0\]

\[2 \bullet \sin{3x} \bullet \cos{4x} - \cos{4x} = 0\]

\[\cos{4x} \bullet \left( 2\sin{3x} - 1 \right) = 0\]

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

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

\[4x = \frac{\pi}{2} + \pi n\]

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

\[x = \frac{\pi}{8} + \frac{\text{πn}}{4}.\]

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

\[2\sin{3x} = 1\]

\[\sin{3x} = \frac{1}{2}\]

\[3x = ( - 1)^{n} \bullet \arcsin\frac{1}{2} + \pi n\]

\[3x = ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n\]

\[x = \frac{1}{3} \bullet \left( ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n \right)\]

\[x = ( - 1)^{n} \bullet \frac{\pi}{18} + \frac{\text{πn}}{3}.\]

\[Ответ:\ \ \frac{\pi}{8} + \frac{\text{πn}}{4};\ \ \]

\[( - 1)^{n} \bullet \frac{\pi}{18} + \frac{\text{πn}}{3}.\]

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

\[2 \bullet \cos\frac{x + 3x}{2} \bullet \cos\frac{x - 3x}{2} = 4\cos{2x}\]

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

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

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

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

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

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

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

\[2)\ \cos x - 2 = 0\]

\[\cos x = 2\]

\[корней\ нет.\]

\[Ответ:\ \ \frac{\pi}{4} + \frac{\text{πn}}{2}.\]

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

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

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

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

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

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

\[y = \cos{2x}:\]

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

\[D = 1 + 8 = 9\]

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

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

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

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

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

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

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

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

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

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

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

\[x = \pm \frac{\pi}{6} + \pi n.\]

\[Ответ:\ \ \frac{\pi}{2} + \pi n;\ \ \pm \frac{\pi}{6} + \pi n.\]