Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 535

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

\[\sin x = 1\]

\[x = \arcsin 1 + 2\pi n\]

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

\[Ответ:\ \ \frac{\pi}{2} + 2\pi n.\]

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

\[- \cos x = 1\]

\[\cos x = - 1\]

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

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

\[Ответ:\ \ \pi + 2\pi n.\]

\[3)\cos(x - \pi) = 0\]

\[\cos\left( - 2\pi + (\pi + x) \right) = 0\]

\[\cos(\pi + x) = 0\]

\[- \cos x = 0\]

\[\cos x = 0\]

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

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

\[Ответ:\ \ \frac{\pi}{2} + \pi n.\]

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

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

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

\[- \cos x = 1\]

\[\cos x = - 1\]

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

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

\[Ответ:\ \ \pi + 2\pi n.\]

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

\[- \sin x = - 1\]

\[\sin x = 1\]

\[x = \arcsin 1 + 2\pi n\]

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

\[Ответ:\ \ \frac{\pi}{2} + 2\pi n.\]

\[\cos{5x} \bullet \cos{2x} + \sin{5x} \bullet \sin{2x} =\]

\[= 0\]

\[\cos{3x} = 0\]

\[3x = \arccos 0 + \pi n = \frac{\pi}{2} + \pi n\]

\[x = \frac{\pi}{6} + \frac{\text{πn}}{3}\]

\[Ответ:\ \ \frac{\pi}{6} + \frac{\text{πn}}{3}\text{.\ }\]