Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 681

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

\[\cos x - \sin x - \frac{1}{\cos x} = 0\ \ \ \ \ |\ :\cos x\]

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

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

\[- tg\ x - tg^{2}\ x = 0\]

\[Пусть\ y = tg\ x,\ тогда:\]

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

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

\[y(y + 1) = 0\]

\[y_{1} = 0\ \ и\ \ y_{2} = - 1.\]

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

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

\[2)\ tg\ x = 0\]

\[x = arctg\ 0 + \pi n = \pi n.\]

\[3)\ tg\ x = - 1\]

\[x = - arctg\ 1 + \pi n\]

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

\[Ответ:\ \ \pi n;\ \ - \frac{\pi}{4} + \pi n.\]

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

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

\[1)\ \cos x + \sin x = 0\ \ \ \ \ |\ :\cos x\]

\[1 + tg\ x = 0\]

\[tg\ x = - 1\]

\[x = - arctg\ 1 + \pi n\]

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

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

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

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

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

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

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

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