Страница 197 номер 656, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

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

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

3) \(\sqrt{2}\cos\left( 2x + \frac{\pi}{4} \right) + 1 = 0\)
\(\cos\left( 2x + \dfrac{\pi}{4} \right) = - \dfrac{\sqrt{2}}{2};\)
\(2x + \dfrac{\pi}{4} = \pm \dfrac{3\pi}{4} + 2\pi n,\)
\(2x = \frac{\pi}{2} + 2\pi n\) или \(2x = - \pi + 2\pi n\)
\(x = \frac{\pi}{4} + \pi n\) или \(x = - \frac{\pi}{2} + \pi n\), \(n \in Z\);

4) \(2\cos\left( \frac{\pi}{3} - 3x \right) - \sqrt{3} = 0\)
\(\cos\left( \dfrac{\pi}{3} - 3x \right) = \dfrac{\sqrt{3}}{2},\)
\(\dfrac{\pi}{3} - 3x = \pm \dfrac{\pi}{6} + 2\pi n,\)
\(3x = \dfrac{\pi}{3} \pm \dfrac{\pi}{6} + 2\pi n,\)
\(x = \dfrac{\pi}{9} \pm \dfrac{\pi}{18} + \dfrac{2\pi}{3}n, n \in Z.\)