Страница 411 номер 1379, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(\sin{5x} = \sin{3x}\)
\(\sin{5x} - \sin{3x} = 0,\)
\(2\sin x\cos{4x} = 0,\)
\(\sin x = 0\) или \(\cos{4x} = 0\)
\(x_{1} = \pi n,\)
\(x_{2} = \dfrac{\pi}{8} + \dfrac{\pi}{4}n, n \in Z;\)

2) \(\cos{6x} + \cos{2x} = 0\)
\(2\cos{4x}\cos{2x} = 0,\)
\(\cos{4x} = 0\) или \(\cos{2x} = 0\)
\(x_{1} = \dfrac{\pi}{8} + \dfrac{\pi}{4}n,\)
\(x_{2} = \dfrac{\pi}{4} + \dfrac{\pi}{2}n, n \in Z;\)

3) \(\sin{3x} + \cos{7x} = 0\)
\(\sin{3x} + \sin\left( \dfrac{\pi}{2} + 7x \right) = 0,\)
\(2\sin\left( \dfrac{\pi}{4} + 5x \right)\cos\left( \dfrac{\pi}{4} + 2x \right) = 0,\)
\(\sin\left( \frac{\pi}{4} + 5x \right) = 0\) или \(\cos\left( \frac{\pi}{4} + 2x \right) = 0\)
\(\frac{\pi}{4} + 5x = \pi n\) или \(\frac{\pi}{4} + 2x = \frac{\pi}{2} + \pi n\)
\(x_{1} = - \dfrac{\pi}{20} + \dfrac{\pi}{5}n,\)
\(x_{2} = \dfrac{\pi}{8} + \dfrac{\pi}{2}n, n \in Z;\)

4) \(\sin x = \cos{5x}\)
\(\sin x - \sin\left( \dfrac{\pi}{2} - 5x \right) = 0,\)
\(2\sin\left( 3x - \dfrac{\pi}{4} \right)\cos\left( \dfrac{\pi}{4} - 2x \right) = 0,\)
\(\sin\left( 3x - \frac{\pi}{4} \right) = 0\) или \(\cos\left( \frac{\pi}{4} - 2x \right) = 0\)
\(3x - \frac{\pi}{4} = \pi n\) или \(\frac{\pi}{4} - 2x = \frac{\pi}{2} + \pi n\)
\(x_{1} = \dfrac{\pi}{12} + \dfrac{\pi}{3}n,\)
\(x_{2} = - \dfrac{\pi}{8} + \dfrac{\pi}{2}n, n \in Z.\)