Страница 411 номер 1366, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(\sin{2x} = 3\sin x\cos^{2}x\)
\(2\sin x\cos x - 3\sin x\cos^{2}x = 0,\)
\(\sin x\cos x\left( 2 - 3\cos x \right) = 0,\)
\(\sin x = 0\) или \(\cos x = 0\) или \(\cos x = \frac{2}{3}\)
\(x_{1} = \pi n, n \in Z\)
\(x_{2} = \dfrac{\pi}{2} + \pi n, n \in Z\)
\(x_{3} = \pm \arccos\dfrac{2}{3} + 2\pi n, n \in Z;\)

2) \(\sin{4x} = \sin{2x}\)
\(2\sin{2x}\cos{2x} - \sin{2x} = 0,\)
\(\sin{2x}\left( 2\cos{2x} - 1 \right) = 0,\)
\(\sin{2x} = 0\) или \(\cos{2x} = \frac{1}{2}\)
\(x_{1} = \dfrac{\pi}{2}n, n \in Z\)
\(x_{2} = \pm \dfrac{\pi}{6} + \pi n, n \in Z;\)

3) \(\cos{2x} + \cos^{2}x = 0\)
\(\cos^{2}x - \sin^{2}x + \cos^{2}x = 0,\)
\(2\cos^{2}x - 1 + \cos^{2}x = 0,\)
\(3\cos^{2}x = 1,\)
\(\cos^{2}x = \dfrac{1}{3},\)
\(\cos x = \pm \dfrac{1}{\sqrt{3}},\)
\(x_{1} = \pm \arccos\dfrac{1}{\sqrt{3}} + 2\pi n, n \in Z\)
\(x_{2} = \pm \left( \pi - \arccos\dfrac{1}{\sqrt{3}} \right) + 2\pi n, n \in Z;\)

4) \(\sin{2x} = \cos^{2}x\)
\(2\sin x\cos x - \cos^{2}x = 0,\)
\(\cos x\left( 2\sin x - \cos x \right) = 0,\)
\(\cos x = 0\) или \(2\operatorname{tg} x - 1 = 0\)
\(x_{1} = \dfrac{\pi}{2} + \pi n, n \in Z\)
\(x_{2} = \operatorname{arctg} \dfrac{1}{2} + \pi n, n \in Z.\)