Страница 192 номер 620, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(\sin^{2}x = \frac{1}{4}\)
\(\sin x = \frac{1}{2}\) или \(\sin x = - \frac{1}{2}\)
\(x_{1} = (-1)^{n} \cdot \dfrac{\pi}{6} + \pi n,\)
\(x_{2} = (-1)^{n + 1} \cdot \dfrac{\pi}{6} + \pi n, n \in Z.\)
\(x = \pm \dfrac{\pi}{6} + \pi n, n \in Z;\)

2) \(\cos^{2}x = \frac{1}{2}\)
\(\cos x = \frac{1}{\sqrt{2}}\) или \(\cos x = - \frac{1}{\sqrt{2}}\)
\(x_{1} = \pm \dfrac{\pi}{4} + 2\pi n,\)
\(x_{2} = \pm \dfrac{3\pi}{4} + 2\pi n, n \in Z.\)
\(x = \dfrac{\pi}{4} + \dfrac{\pi}{2}n, n \in Z;\)

3) \(2\sin^{2}x + \sin x - 1 = 0\)
\(\sin x = z, 2z^{2} + z - 1 = 0,\)
\(z_{1{,}2} = \dfrac{-1 \pm \sqrt{1 + 8}}{4},\)
\(z_{1} = \dfrac{1}{2}, z_{2} = - 1\)
\(\sin x = \frac{1}{2}\) или \(\sin x = - 1\)
\(x_{1} = (-1)^{n} \cdot \dfrac{\pi}{6} + \pi n,\)
\(x_{2} = - \dfrac{\pi}{2} + 2\pi n, n \in Z;\)

4) \(2\cos^{2}x + \cos x - 6 = 0\)
\(\cos x = z, 2z^{2} + z - 6 = 0,\)
\(z_{1{,}2} = \dfrac{-1 \pm \sqrt{1 + 48}}{4},\)
\(z_{1} = \dfrac{3}{2}, z_{2} = - 2,\)
\(\cos x = \frac{3}{2}\) или \(\cos x = - 2\). Корней нет.