Страница 172 номер 573, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(\cos{4x} = 1\)
\(4x = 2\pi k, k \in Z,\)
\(x = \dfrac{\pi}{2}k, k \in Z;\)

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

3) \(\sqrt{2}\cos\frac{x}{4} = - 1\)
\(\dfrac{x}{4} = \pm \left( \pi - \arccos\dfrac{\sqrt{2}}{2} \right) + 2\pi k,\)
\(\dfrac{x}{4} = \pm \dfrac{3\pi}{4} + 2\pi k,\)
\(x = \pm 3\pi + 8\pi k, k \in Z;\)

4) \(2\cos\frac{x}{3} = \sqrt{3}\)
\(\dfrac{x}{3} = \pm \arccos\dfrac{\sqrt{3}}{2} + 2\pi k,\)
\(\dfrac{x}{3} = \pm \dfrac{\pi}{6} + 2\pi k,\)
\(x = \pm \dfrac{\pi}{2} + 6\pi k, k \in Z;\)

5) \(\cos\left( x + \frac{\pi}{3} \right) = 0\)
\(x + \dfrac{\pi}{3} = \dfrac{\pi}{2} + \pi k,\)
\(x = \dfrac{\pi}{6} + \pi k, k \in Z;\)

6) \(\cos\left( 2x - \frac{\pi}{4} \right) = 0\)
\(2x - \dfrac{\pi}{4} = \dfrac{\pi}{2} + \pi k,\)
\(2x = \dfrac{3\pi}{4} + \pi k,\)
\(x = \dfrac{3\pi}{8} + \dfrac{\pi}{2}k, k \in Z.\)