Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 596

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\[1)\ \left( 4\sin x - 3 \right)\left( 2\sin x + 1 \right) = 0\]

\[1)\ 4\sin x - 3 = 0\]

\[4\sin x = 3\]

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

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

\[2)\ 2\sin x + 1 = 0\]

\[\sin x = - 1\]

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

\[x = ( - 1)^{n + 1} \bullet \arcsin\frac{1}{2} + \pi n\]

\[x = ( - 1)^{n + 1} \bullet \frac{\pi}{6} + \pi n\]

\[Ответ:\ \ \]

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

\[x = \ ( - 1)^{n + 1} \bullet \frac{\pi}{6} + \pi n.\]

\[2)\ \left( 4\sin{3x} - 1 \right)\left( 2\sin x + 3 \right) = 0\]

\[1)\ 4\sin{3x} - 1 = 0\]

\[4\sin{3x} = 1\]

\[\sin{3x} = \frac{1}{4}\]

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

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

\[2)\ 2\sin x + 3 = 0\]

\[2\sin x = - 3\]

\[\sin x = - \frac{3}{2}\]

\[корней\ нет\]

\[Ответ:\ \]

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