Страница 193 номер 632, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

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

2) \(\sqrt{2}\cos\left( x - \frac{\pi}{4} \right) = \left( \sin x + \cos x \right)^{2}\)
\(\sqrt{2}\left( \dfrac{\sqrt{2}}{2}\cos x + \dfrac{\sqrt{2}}{2}\sin x \right) = \left( \sin x + \cos x \right)^{2},\)
\(\left( \sin x + \cos x \right)\left( 1 - \sin x - \cos x \right) = 0,\)
\(\sin x + \cos x = 0\) или \(\sin x + \cos x = 1\)
\(\operatorname{tg} x = - 1\) или \(\sin\left( x + \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2}\)
\(x_{1} = - \dfrac{\pi}{4} + \pi n,\)
\(x_{2} = (-1)^{n} \cdot \dfrac{\pi}{4} - \dfrac{\pi}{4} + \pi n, n \in Z.\)