Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1544

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\[1)\ f(x) = \sin{2x} - x;\]

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

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

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

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

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

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

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

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

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

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

\[2)\ f(x) = \cos{2x} + 2x;\]

\[f^{'}(x) = \left( \cos{2x} \right)^{'} + (2x)^{'} =\]

\[= - 2\sin{2x} + 2;\]

\[- 2\sin{2x} + 2 = 0;\]

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

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

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

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

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

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

\[Ответ:\ \ \frac{\pi}{4} + \pi n.\]