Страница 141 номер 474, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Решить уравнение: 1) \(2\sin x + \sin^2 x + \cos^2 x = 1\); 2) \(2\sin^2 x + 3\cos^2 x - 2 = 0\); 3) \(3\cos^2 x - 2\sin x = 3 - 3\sin^2 x\); 4) \(\cos^2 x - \sin^2 x = 2\sin x - 1 - 2\sin^2 x\).
1) \(2\sin x + \sin^{2} x + \cos^{2} x = 1\)
\(2\sin x + 1 = 1,\)
\(2\sin x = 0,\)
\(\sin x = 0,\)
\(x = \pi k,\ k \in Z;\)
2) \(2\sin^{2} x + 3\cos^{2} x - 2 = 0\)
\(2(\sin^{2} x + \cos^{2} x) + \cos^{2} x - 2 = 0,\)
\(2 - 2 + \cos^{2} x = 0,\)
\(\cos^{2} x = 0,\)
\(x = \dfrac{\pi}{2} + \pi k,\ k \in Z;\)
3) \(3\cos^{2} x - 2\sin x = 3 - 3\sin^{2} x\)
\(3(\cos^{2} x + \sin^{2} x) - 2\sin x - 3 = 0,\)
\(3 - 3 - 2\sin x = 0,\)
\(\sin x = 0,\)
\(x = \pi k,\ k \in Z;\)
4) \(\cos^{2} x - \sin^{2} x = 2\sin x - 1 - 2\sin^{2} x\)
\(\cos^{2} x - \sin^{2} x + 2\sin^{2} x - 2\sin x + 1 = 0,\)
\(1 - 2\sin x + 1 = 0,\)
\(2\sin x = 2,\)
\(\sin x = 1,\)
\(x = \dfrac{\pi}{2} + 2\pi k,\ k \in Z.\)
