Страница 192 номер 630, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

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

2) \(2\cos^{2}{2x} - 1 = \sin{4x}\)
\(2\cos^{2}{2x} - \cos^{2}{2x} - \sin^{2}{2x} = \sin{4x},\)
\(\cos{4x} = \sin{4x},\)
\(\operatorname{tg} {4x} = 1,\)
\(x = \dfrac{\pi}{16} + \dfrac{\pi}{4}n, n \in Z;\)

3) \(2\cos^{2}{2x} + 3\cos^{2}x = 2\)
\(2\cos^{2}{2x} + \dfrac{3}{2}\left( 2 - 2\sin^{2}x \right) = 2,\)
\(2\cos^{2}{2x} + \dfrac{3}{2}\left( 1 + \cos^{2}x - \sin^{2}x \right) = 2,\)
\(2\cos^{2}{2x} + \dfrac{3}{2}\left( 1 + \cos{2x} \right) - 2 = 0,\)
\(4\cos^{2}{2x} + 3\cos{2x} - 1 = 0,\)
\(\cos{2x} = z, 4z^{2} + 3z - 1 = 0,\)
\(z_{1{,}2} = \dfrac{-3 \pm \sqrt{9 + 16}}{8},\)
\(z_{1} = \dfrac{1}{4}, z_{2} = - 1\)
\(\cos{2x} = \frac{1}{4}\) или \(\cos{2x} = - 1\)
\(x_{1} = \pm \dfrac{1}{2}\arccos\dfrac{1}{4} + \pi n,\)
\(x_{2} = \dfrac{\pi}{2} + \pi n, n \in Z;\)

4) \(\left( \sin x + \cos x \right)^{2} = 1 + \cos x\)
\(1 + 2\sin x\cos x - 1 - \cos x = 0,\)
\(\cos x\left( 2\sin x - 1 \right) = 0,\)
\(\cos x = 0\) или \(\sin x = \frac{1}{2}\)
\(x_{1} = \dfrac{\pi}{2} + \pi n,\)
\(x_{2} = (-1)^{n} \cdot \dfrac{\pi}{6} + \pi n, n \in Z.\)