Страница 143 номер 480, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(\sin(-x) = 1\)
\(-\sin x = 1,\)
\(\sin x = -1,\)
\(x = \dfrac{3\pi}{2} + 2\pi k,\ k \in Z;\)

2) \(\cos(-2x) = 0\)
\(\cos 2x = 0,\)
\(2x = \dfrac{\pi}{2} + \pi k,\ k \in Z,\)
\(x = \dfrac{\pi}{4} + \dfrac{\pi}{2}k,\ k \in Z;\)

3) \(\cos(-2x) = 1\)
\(\cos 2x = 1,\)
\(2x = 2\pi k,\ k \in Z,\)
\(x = \pi k,\ k \in Z;\)

4) \(\sin(-2x) = 0\)
\(-\sin 2x = 0,\)
\(2x = \pi k,\ k \in Z,\)
\(x = \dfrac{\pi}{2}k,\ k \in Z;\)

5) \(\cos^{2}(-x) + \sin(-x) = 2 - \sin^{2}x\)
\(\cos^{2}x - \sin x - 2 + \sin^{2}x = 0,\)
\(1 - 2 - \sin x = 0,\)
\(\sin x = -1,\)
\(x = \dfrac{3\pi}{2} + 2\pi k,\ k \in Z;\)

6) \(1 - \sin^{2}(-x) + \cos(4\pi - x) = \cos(x - 2\pi)\)
\(1 - \sin^{2}x + \cos x - \cos x = 0,\)
\(\cos^{2}x = 0,\)
\(x = \dfrac{\pi}{2} + \pi k,\ k \in Z.\)