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МатематикаГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1091
Задание 1091
\[\boxed{\mathbf{1091}\mathbf{.}}\]
\[1)\cos\left( \frac{\pi}{2} - x \right) = 1\]
\[\sin x = 1\]
\[x = \arcsin 1 + 2\pi n\]
\[x = \frac{\pi}{2} + 2\pi n\]
\[Ответ:\ \ \frac{\pi}{2} + 2\pi n.\]
\[2)\sin\left( \frac{3\pi}{2} + x \right) = 1\]
\[- \cos x = 1\]
\[\cos x = - 1\]
\[x = \arccos( - 1) + 2\pi n\]
\[x = \pi + 2\pi n\]
\[Ответ:\ \ \pi + 2\pi n.\]
\[3)\cos(x - \pi) = 0\]
\[\cos\left( - 2\pi + (\pi + x) \right) = 0\]
\[\cos(\pi + x) = 0\]
\[- \cos x = 0\]
\[\cos x = 0\]
\[x = \arccos 0 + \pi n\]
\[x = \frac{\pi}{2} + \pi n\]
\[Ответ:\ \ \frac{\pi}{2} + \pi n.\]
\[4)\sin\left( x - \frac{\pi}{2} \right) = 1\]
\[\sin\left( - 2\pi + \left( \frac{3\pi}{2} + x \right) \right) = 1\]
\[\sin\left( \frac{3\pi}{2} + x \right) = 1\]
\[- \cos x = 1\]
\[\cos x = - 1\]
\[x = \arccos( - 1) + 2\pi n\]
\[x = \pi + 2\pi n\]
\[Ответ:\ \ \pi + 2\pi n.\]
\[5)\sin{(2x + 3\pi}) \bullet \sin\left( 3x + \frac{3\pi}{2} \right) -\]
\[- \sin{3x} \bullet \cos{2x} = - 1\]
\[\sin(\pi + 2x) \bullet \left( - \cos{3x} \right) -\]
\[- \sin{3x} \bullet \cos{2x} = - 1\]
\[\sin{2x} \bullet \cos{3x} - \sin{3x} \bullet \cos{2x} =\]
\[= - 1\]
\[\frac{1}{2}\left( \sin(2x + 3x) + \sin(2x - 3x) \right) -\]
\[- \frac{1}{2}\left( \sin(3x + 2x) + \sin(3x - 2x) \right) = - 1\]
\[\frac{\sin{5x} + \sin( - x)}{2} -\]
\[- \frac{\sin{5x} + \sin x}{2} = - 1\]
\[\frac{- \sin x}{2} - \frac{\sin x}{2} = - 1\]
\[- \sin x = - 1\]
\[\sin x = 1\]
\[x = \arcsin 1 + 2\pi n\]
\[x = \frac{\pi}{2} + 2\pi n\]
\[Ответ:\ \ \frac{\pi}{2} + 2\pi n.\]
\[6)\sin\left( 5x - \frac{3\pi}{2} \right) \bullet\]
\[\bullet \cos(2x + 4\pi) - \sin(5x + \pi) \bullet\]
\[\bullet \sin{2x} = 0\]
\[\sin\left( - 2\pi + \left( \frac{\pi}{2} + 5x \right) \right) \bullet \cos{2x} +\]
\[+ \sin{5x} \bullet \sin{2x} = 0\]
\[\sin\left( \frac{\pi}{2} + 5x \right) \bullet \cos{2x} +\]
\[+ \sin{5x} \bullet \sin{2x} = 0\]
\[\cos{5x} \bullet \cos{2x} +\]
\[+ \sin{5x} \bullet \sin{2x} = 0\]
\[\frac{1}{2}\left( \cos(5x + 2x) + \cos(5x - 2x) \right) +\]
\[+ \frac{1}{2}\left( \cos(5x - 2x) - \cos(5x + 2x) \right) = 0\]
\[\frac{\cos{7x} + \cos{3x}}{2} +\]
\[+ \frac{\cos{3x} - \cos{7x}}{2} = 0\]
\[\cos{3x} = 0\]
\[3x = \arccos 0 + \pi n = \frac{\pi}{2} + \pi n\]
\[x = \frac{\pi}{6} + \frac{\text{πn}}{3}\]
\[Ответ:\ \ \frac{\pi}{6} + \frac{\text{πn}}{3}\text{.\ }\]
