Страница 411 номер 1373, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(\sin x + \sin{2x} = \cos x + 2\cos^{2}x\)
\(\sin x\left( 1 + 2\cos x \right) = \cos x\left( 1 + 2\cos x \right),\)
\(\left( 1 + 2\cos x \right)\left( \sin x - \cos x \right) = 0,\)
\(\cos x = - \frac{1}{2}\) или \(\operatorname{tg} x = 1\)
\(x_{1} = \pm \left( \pi - \dfrac{\pi}{3} \right) + 2\pi n,\)
\(x_{1} = \pm \dfrac{2\pi}{3} + 2\pi n,\)
\(x_{2} = \dfrac{\pi}{4} + \pi n, n \in Z;\)

2) \(2\cos{2x} = \sqrt{6}\left( \cos x - \sin x \right)\)
\(2\left( \cos x - \sin x \right)\left( \cos x + \sin x \right) = \sqrt{6}\left( \cos x - \sin x \right),\)
\(\left( \cos x - \sin x \right)\left( 2\cos x + 2\sin x - \sqrt{6} \right) = 0,\)
\(\operatorname{tg} x = 1\) или \(\cos x + \sin x = \frac{\sqrt{3}}{\sqrt{2}}\)
\(x_{1} = \dfrac{\pi}{4} + \pi n, n \in Z;\)
\(\sin\left( \dfrac{\pi}{2} - x \right) + \sin x = \dfrac{\sqrt{3}}{\sqrt{2}},\)
\(2\sin\dfrac{\pi}{4}\cos\left( \dfrac{\pi}{4} - x \right) = \dfrac{\sqrt{3}}{\sqrt{2}},\)
\(\cos\left( \dfrac{\pi}{4} - x \right) = \dfrac{\sqrt{3}}{2},\)
\(\dfrac{\pi}{4} - x = \pm \dfrac{\pi}{6} + 2\pi n,\)
\(x_{2} = \dfrac{\pi}{12} + 2\pi n,\)
\(x_{3} = \dfrac{5\pi}{12} + 2\pi n, n \in Z.\)