Страница 193 номер 636, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Решить уравнение (620—644). 1) \(4 \sin^2 x - 5 \sin x \cos x - 6 \cos^2 x = 0\); 2) \(3 \sin^2 x - 7 \sin x \cos x + 2 \cos^2 x = 0\); 3) \(1 - 4 \sin x \cos x + 4 \cos^2 x = 0\); 4) \(1 + \sin^2 x = 2 \sin x \cos x\).
1) \(4\sin^{2}x - 5\sin x\cos x - 6\cos^{2}x = 0\)
\(4\operatorname{tg}^{2}x - 5\operatorname{tg} x - 6 = 0,\)
\(\operatorname{tg} x = z, 4z^{2} - 5z - 6 = 0,\)
\(z_{1{,}2} = \dfrac{5 \pm \sqrt{25 + 96}}{8},\)
\(z_{1} = 2, z_{2} = - \dfrac{3}{4}\)
\(\operatorname{tg} x = 2\) или \(\operatorname{tg} x = - \frac{3}{4}\)
\(x_{1} = \operatorname{arctg} 2 + \pi n,\)
\(x_{2} = - \operatorname{arctg} \dfrac{3}{4} + \pi n, n \in Z;\)
2) \(3\sin^{2}x - 7\sin x\cos x + 2\cos^{2}x = 0\)
\(3\operatorname{tg}^{2}x - 7\operatorname{tg} x + 2 = 0,\)
\(\operatorname{tg} x = z, 3z^{2} - 7z + 2 = 0,\)
\(z_{1{,}2} = \dfrac{7 \pm \sqrt{49 - 24}}{6},\)
\(z_{1} = 2, z_{2} = \dfrac{1}{3}\)
\(\operatorname{tg} x = 2\) или \(\operatorname{tg} x = \frac{1}{3}\)
\(x_{1} = \operatorname{arctg} 2 + \pi n,\)
\(x_{2} = \operatorname{arctg} \dfrac{1}{3} + \pi n, n \in Z;\)
3) \(1 - 4\sin x\cos x + 4\cos^{2}x = 0\)
\(\sin^{2}x - 4\sin x\cos x + 5\cos^{2}x = 0,\)
\(\operatorname{tg}^{2}x - 4\operatorname{tg} x + 5 = 0,\)
\(\operatorname{tg} x = z, z^{2} - 4z + 5 = 0,\)
\(D < 0\) – корней нет;
4) \(1 + \sin^{2}x = 2\sin x\cos x\)
\(\cos^{2}x + 2\sin^{2}x - 2\sin x\cos x = 0,\)
\(1 + 2\operatorname{tg}^{2}x - 2\operatorname{tg} x = 0,\)
\(\operatorname{tg} x = z, 2z^{2} - 2z + 1 = 0,\)
\(D < 0\) – корней нет.
