Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 663

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\[1)\ 2\sin{2x} = 3\cos{2x}\ \ \ |\ :\cos{2x}\]

\[2\ tg\ 2x = 3\]

\[tg\ 2x = \frac{3}{2}\]

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

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

\[x = \frac{1}{2}\text{arctg}\frac{3}{2} + \frac{\text{πn}}{2}.\]

\[Ответ:\ \ \frac{1}{2}\text{arctg}\frac{3}{2} + \frac{\text{πn}}{2}.\]

\[2)\ 4\sin{3x} + 5\cos{3x} = 0\text{\ \ }\ \ |\ :\cos{3x}\]

\[4\ tg\ 3x + 5 = 0\]

\[4\ tg\ 3x = - 5\]

\[tg\ 3x = - \frac{5}{4}\]

\[3x = - arctg\frac{5}{4} + \pi n\]

\[x = \frac{1}{3} \bullet \left( - arctg\frac{5}{4} + \pi n \right)\]

\[x = - \frac{1}{3}\text{arctg}\frac{5}{4} + \frac{\text{πn}}{3}.\]

\[Ответ:\ - \frac{1}{3}\text{arctg}\frac{5}{4} + \frac{\text{πn}}{3}.\]