Страница 192 номер 621, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(2\cos^{2}x - \sin x + 1 = 0\)
\(2\left( 1 - \sin^{2}x \right) - \sin x + 1 = 0,\)
\(2\sin^{2}x + \sin x - 3 = 0,\)
\(\sin x = z, 2z^{2} + z - 3 = 0,\)
\(z_{1{,}2} = \dfrac{-1 \pm \sqrt{1 + 24}}{4},\)
\(z_{1} = 1, z_{2} = - \dfrac{3}{2}\)
\(\sin x = 1\) или \(\sin x = - \frac{3}{2}\) – корней нет.
\(x = \dfrac{\pi}{2} + 2\pi n, n \in Z;\)

2) \(3\cos^{2}x - \sin x - 1 = 0\)
\(3\left( 1 - \sin^{2}x \right) - \sin x - 1 = 0,\)
\(3\sin^{2}x + \sin x - 2 = 0,\)
\(\sin x = z, 3z^{2} + z - 2 = 0,\)
\(z_{1{,}2} = \dfrac{-1 \pm \sqrt{1 + 24}}{6},\)
\(z_{1} = \dfrac{2}{3}, z_{2} = - 1\)
\(\sin x = \frac{2}{3}\) или \(\sin x = - 1\)
\(x_{1} = (-1)^{n}\arcsin\dfrac{2}{3} + \pi n,\)
\(x_{2} = - \dfrac{\pi}{2} + 2\pi n, n \in Z;\)

3) \(4\sin^{2}x - \cos x - 1 = 0\)
\(4\left( 1 - \cos^{2}x \right) - \cos x - 1 = 0,\)
\(4\cos^{2}x + \cos x - 3 = 0,\)
\(\cos x = z, 4z^{2} + z - 3 = 0,\)
\(z_{1{,}2} = \dfrac{-1 \pm \sqrt{1 + 48}}{8},\)
\(z_{1} = \dfrac{3}{4}, z_{2} = - 1\)
\(\cos x = \frac{3}{4}\) или \(\cos x = - 1\)
\(x_{1} = \pm \arccos\dfrac{3}{4} + 2\pi n,\)
\(x_{2} = \pi + 2\pi n, n \in Z;\)

4) \(2\sin^{2}x + 3\cos x = 0\)
\(2\left( 1 - \cos^{2}x \right) + 3\cos x = 0,\)
\(2\cos^{2}x - 3\cos x - 2 = 0,\)
\(\cos x = z, 2z^{2} - 3z - 2 = 0\)
\(z_{1{,}2} = \dfrac{3 \pm \sqrt{9 + 16}}{4},\)
\(z_{1} = 2, z_{2} = - \dfrac{1}{2}\)
\(\cos x = 2\) – корней нет;
\(\cos x = - \dfrac{1}{2},\)
\(x = \pm \dfrac{2\pi}{3} + 2\pi n, n \in Z.\)