Страница 192 номер 625, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(\sin x - \cos x = 1\)
\(\dfrac{\sqrt{2}}{2}\sin x - \dfrac{\sqrt{2}}{2}\cos x = \dfrac{\sqrt{2}}{2},\)
\(\sin x\cos\dfrac{\pi}{4} - \cos x\sin\dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2},\)
\(\sin\left( x - \dfrac{\pi}{4} \right) = \dfrac{\sqrt{2}}{2},\)
\(x - \dfrac{\pi}{4} = (-1)^{n} \cdot \dfrac{\pi}{4} + \pi n,\)
\(x = (-1)^{n} \cdot \dfrac{\pi}{4} + \dfrac{\pi}{4} + \pi n, n \in Z;\)

2) \(\sin x + \cos x = 1\)
\(\dfrac{\sqrt{2}}{2}\sin x + \dfrac{\sqrt{2}}{2}\cos x = \dfrac{\sqrt{2}}{2},\)
\(\sin x\cos\dfrac{\pi}{4} + \cos x\sin\dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2},\)
\(\sin\left( x + \dfrac{\pi}{4} \right) = \dfrac{\sqrt{2}}{2},\)
\(x + \dfrac{\pi}{4} = (-1)^{n} \cdot \dfrac{\pi}{4} + \pi n,\)
\(x = (-1)^{n} \cdot \dfrac{\pi}{4} - \dfrac{\pi}{4} + \pi n, n \in Z;\)

3) \(\sqrt{3}\sin x + \cos x = 2\)
\(\dfrac{\sqrt{3}}{2}\sin x + \dfrac{1}{2}\cos x = 1,\)
\(\sin x\cos\dfrac{\pi}{6} + \cos x\sin\dfrac{\pi}{6} = 1,\)
\(\sin\left( x + \dfrac{\pi}{6} \right) = 1,\)
\(x + \dfrac{\pi}{6} = \dfrac{\pi}{2} + 2\pi n,\)
\(x = \dfrac{\pi}{3} + 2\pi n, n \in Z;\)

4) \(\sin{3x} + \cos{3x} = \sqrt{2}\)
\(\dfrac{\sqrt{2}}{2}\sin{3x} + \dfrac{\sqrt{2}}{2}\cos{3x} = 1,\)
\(\sin{3x}\cos\dfrac{\pi}{4} + \cos{3x}\sin\dfrac{\pi}{4} = 1,\)
\(\sin\left( 3x + \dfrac{\pi}{4} \right) = 1,\)
\(3x + \dfrac{\pi}{4} = \dfrac{\pi}{2} + 2\pi n,\)
\(3x = \dfrac{\pi}{4} + 2\pi n,\)
\(x = \dfrac{\pi}{12} + \dfrac{2\pi}{3}n, n \in Z.\)