Страница 411 номер 1378, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Решить уравнение (1363—1385). 1) \(4\sin^2 x - 8\sin x \cos x + 10\cos^2 x = 3\); 2) \(3\sin^2 x - 2\sin x \cos x = 1\).
1) \(4\sin^{2}x - 8\sin x\cos x + 10\cos^{2}x = 3\)
\(4\sin^{2}x - 3\sin^{2}x - 3\cos^{2}x - 8\sin x\cos x + 10\cos^{2}x = 0,\)
\(\sin^{2}x + 7\cos^{2}x - 8\sin x\cos x = 0,\)
\(\operatorname{tg}^{2}x - 8\operatorname{tg} x + 7 = 0,\)
\(\operatorname{tg} x = z, z^{2} - 8z + 7 = 0\)
\(z_{1{,}2} = \dfrac{8 \pm \sqrt{64 - 28}}{2},\)
\(z_{1} = 7, z_{2} = 1\)
\(\operatorname{tg} x = 7\) или \(\operatorname{tg} x = 1\)
\(x = \operatorname{arctg} 7 + \pi n,\)
\(x = \dfrac{\pi}{4} + \pi n, n \in Z;\)
2) \(3\sin^{2}x - 2\sin x\cos x = 1\)
\(3\sin^{2}x - \sin^{2}x - \cos^{2}x - 2\sin x\cos x = 0,\)
\(2\sin^{2}x - \cos^{2}x - 2\sin x\cos x = 0,\)
\(2\operatorname{tg}^{2}x - 2\operatorname{tg} x - 1 = 0,\)
\(\operatorname{tg} x = z, 2z^{2} - 2z - 1 = 0\)
\(z_{1{,}2} = \dfrac{2 \pm \sqrt{4 + 8}}{4} = \dfrac{1 \pm \sqrt{3}}{2},\)
\(\operatorname{tg} x = \dfrac{1 \pm \sqrt{3}}{2},\)
\(x = \operatorname{arctg} \dfrac{1 \pm \sqrt{3}}{2} + \pi n, n \in Z.\)
