Страница 148 номер 497, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(\cos 6x \cos 5x + \sin 6x \sin 5x = -1\)
\(\cos(6x - 5x) = -1,\)
\(\cos x = -1,\)
\(x = \pi + 2\pi k,\ k \in Z;\)

2) \(\sin 3x \cos 5x - \sin 5x \cos 3x = -1\)
\(\sin(3x - 5x) = -1,\)
\(\sin(-2x) = -1,\)
\(\sin 2x = 1,\)
\(2x = \dfrac{\pi}{2} + 2\pi k,\ k \in Z,\)
\(x = \dfrac{\pi}{4} + \pi k,\ k \in Z;\)

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

4) \(\sqrt{2}\sin\left(\frac{\pi}{4} - \frac{x}{2}\right) + \sin\frac{x}{2} = 1\)
\(\sqrt{2}\sin\dfrac{\pi}{4}\cos\dfrac{x}{2} - \sqrt{2}\cos\dfrac{\pi}{4}\sin\dfrac{x}{2} + \sin\dfrac{x}{2} = 1,\)
\(\sqrt{2} \cdot \dfrac{\sqrt{2}}{2}\cos\dfrac{x}{2} - \sqrt{2} \cdot \dfrac{\sqrt{2}}{2}\sin\dfrac{x}{2} + \sin\dfrac{x}{2} = 1,\)
\(\cos\dfrac{x}{2} = 1,\)
\(\dfrac{x}{2} = 2\pi k,\ k \in Z,\)
\(x = 4\pi k,\ k \in Z.\)