Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 624

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\[1)\ \sqrt{3}\cos x + \sin x = 0\ \ \ |\ :\cos x\]

\[\sqrt{3} + tg\ x = 0\]

\[tg\ x = - \sqrt{3}\]

\[x = - arctg\ \sqrt{3} + \pi n\]

\[x = - \frac{\pi}{3} + \pi n.\]

\[Ответ:\ - \frac{\pi}{3} + \pi n.\]

\[2)\cos x = \sin x\ \ \ \ \ |\ :\cos x\ \]

\[1 = tg\ x\]

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

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

\[Ответ:\ \ \frac{\pi}{4} + \pi n.\]

\[3)\sin x = 2\cos x\ \ \ \ \ |\ :\cos x\]

\[tg\ x = 2\]

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

\[Ответ:\ \ arctg\ 2 + \pi n.\]

\[4)\ 2\sin x + \cos x = 0\ \ \ \ \ |\ :\cos x\]

\[2\ tg\ x + 1 = 0\]

\[2\ tg\ x = - 1\]

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

\[x = - arctg\frac{1}{2} + \pi n.\]

\[Ответ:\ - arctg\frac{1}{2} + \pi n.\]