Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1365

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\[1)\ \left( 3 - 4\sin x \right)\left( 3 + 4\cos x \right) = 0\]

\[3 - 4\sin x = 0\]

\[4\sin x = 3\]

\[\sin x = \frac{3}{4}\]

\[x = ( - 1)^{n} \bullet \arcsin\frac{3}{4} + \pi n.\]

\[3 + 4\cos x = 0\]

\[4\cos x = - 3\]

\[\cos x = - \frac{3}{4}\]

\[x = \pm \left( \pi - \arccos\frac{3}{4} \right) + 2\pi n.\]

\[Ответ:\ \ ( - 1)^{n} \bullet \arcsin\frac{3}{4} + \pi n;\ \ \]

\[\ \ \ \ \ \ \ \ \ \ \pm \left( \pi - \arccos\frac{3}{4} \right) + 2\pi n.\]

\[2)\ (tg\ x + 3)(tg\ x + 1) = 0\]

\[tg\ x + 3 = 0\]

\[tg\ x = - 3\]

\[x = - arctg\ 3 + \pi n.\]

\[tg\ x + 1 = 0\]

\[tg\ x = - 1\]

\[x = - arctg\ 1 + \pi n = - \frac{\pi}{4} + \pi n.\]

\[Ответ:\ \ - arctg\ 3 + \pi n;\ \ \]

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