Страница 198 номер 673, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

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

2) \(\sin^{2}x + \cos^{2}{2x} = 1\)
\(\cos^{2}{2x} = \cos^{2}x,\)
\(\left( \cos^{2}x - \sin^{2}x \right)^{2} - \cos^{2}x = 0,\)
\(\cos^{4}x - 2\cos^{2}x\sin^{2}x + \sin^{4}x - \cos^{2}x = 0,\)
\(\cos^{4}x - 2\cos^{2}x + 2\cos^{4}x + 1 - 2\cos^{2}x + \cos^{4}x - \cos^{2}x = 0,\)
\(4\cos^{4}x - 5\cos^{2}x + 1 = 0,\)
\(\cos^{2}x = z,\)
\(4z^{2} - 5z + 1 = 0,\)
\(z_{1{,}2} = \dfrac{5 \pm \sqrt{25 - 16}}{8},\)
\(z_{1} = 1, z_{2} = \dfrac{1}{4}\)
\(\cos^{2}x = 1\) или \(\cos^{2}x = \frac{1}{4}\)
\(\cos x = \pm 1\) или \(\cos x = \pm \frac{1}{2}\)
\(x_{1} = \pi n,\)
\(x_{2} = \pm \dfrac{\pi}{3} + \pi n, n \in Z;\)

3) \(\sin{4x} = 6\cos^{2}{2x} - 4\)
\(2\sin{2x}\cos{2x} = 6\cos^{2}{2x} - 4\cos^{2}{2x} - 4\sin^{2}{2x},\)
\(2\sin{2x}\cos{2x} + 4\sin^{2}{2x} - 2\cos^{2}{2x} = 0,\)
\(4\operatorname{tg}^{2}{2x} + 2\operatorname{tg} {2x} - 2 = 0,\)
\(\operatorname{tg} {2x} = z,\)
\(4z^{2} + 2z - 2 = 0,\)
\(z_{1{,}2} = \dfrac{-2 \pm \sqrt{4 + 32}}{8},\)
\(z_{1} = \dfrac{1}{2}, z_{2} = - 1\)
\(\operatorname{tg} {2x} = \frac{1}{2}\) или \(\operatorname{tg} {2x} = - 1\)
\(x_{1} = \dfrac{1}{2}\operatorname{arctg} \dfrac{1}{2} + \dfrac{\pi}{2}n,\)
\(x_{2} = - \dfrac{\pi}{8} + \dfrac{\pi}{2}n, n \in Z;\)

4) \(2\cos^{2}{3x} + \sin{5x} = 1\)
\(2\cos^{2}{3x} - \cos^{2}{3x} - \sin^{2}{3x} + \sin{5x} = 0,\)
\(\cos{6x} + \sin{5x} = 0,\)
\(\cos{6x} + \cos\left( \dfrac{\pi}{2} - 5x \right) = 0,\)
\(2\cos\left( \dfrac{\pi}{4} + \dfrac{x}{2} \right)\cos\left(-\dfrac{\pi}{4} + \dfrac{11x}{2} \right) = 0,\)
\(\cos\left( \frac{\pi}{4} + \frac{x}{2} \right) = 0\) или \(\cos\left(-\frac{\pi}{4} + \frac{11x}{2} \right) = 0\)
\(\frac{x}{2} + \frac{\pi}{4} = \frac{\pi}{2} + \pi n\) или \(\frac{11x}{2} - \frac{\pi}{4} = \frac{\pi}{2} + \pi n\)
\(\frac{x}{2} = \frac{\pi}{4} + \pi n\) или \(\frac{11x}{2} = \frac{3\pi}{4} + \pi n\)
\(x_{1} = \dfrac{\pi}{2} + 2\pi n,\)
\(x_{2} = \dfrac{3\pi}{22} + \dfrac{2\pi}{11}n, n \in Z.\)