Страница 193 номер 638, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(\sin^{2}x + \sin^{2}{2x} = \sin^{2}{3x}\)
\(\left( \sin x - \sin{3x} \right)\left( \sin x + \sin{3x} \right) + \sin^{2}{2x} = 0,\)
\(2\sin(-x)\cos{2x} \cdot 2\sin{2x}\cos x + \sin^{2}{2x} = 0,\)
\(\sin^{2}{2x}\left( 1 - 2\cos{2x} \right) = 0,\)
\(\sin{2x} = 0\) или \(\cos{2x} = \frac{1}{2}\)
\(x_{1} = \dfrac{\pi}{2}n,\)
\(x_{2} = \pm \dfrac{\pi}{6} + \pi n, n \in Z;\)

2) \(\sin x\left( 1 - \cos x \right)^{2} + \cos x\left( 1 - \sin x \right)^{2} = 2\)
\(\sin x - 2\sin x\cos x + \sin x\cos^{2}x + \cos x - 2\sin x\cos x +\)
\(+ \sin^{2}x\cos x = 2,\)
\(\sin x + \cos x - 4\sin x\cos x + \sin x\cos x\left( \cos x + \sin x \right) = 2,\)
\(\sin x + \cos x + \sin x\cos x\left( \cos x + \sin x \right) = 2\left( \cos^{2}x + \sin^{2}x \right) +\)
\(+ 4\sin x\cos x,\)
\(\sin x + \cos x + \sin x\cos x\left( \cos x + \sin x \right) = 2\left( \cos x + \sin x \right)^{2},\)
\(\sin x + \cos x + \dfrac{\left( \sin x + \cos x \right)^{2} - 1}{2}\left( \cos x + \sin x \right) =\)
\(= 2\left( \cos x + \sin x \right)^{2},\)
\(\left( \cos x + \sin x \right) = z, z + \dfrac{z^{2} - 1}{2}z = 2z^{2},\)
\(4z^{2} - 2z - z^{3} + z = 0,\)
\(z\left( z^{2} - 4z + 1 \right) = 0,\)
\(z_{1} = 0,\)
\(z_{2{,}3} = \dfrac{4 \pm \sqrt{16 - 4}}{2},\)
\(z_{2} = 2 + \sqrt{3}, z_{3} = 2 - \sqrt{3}\)
\(\cos x + \sin x = 0\) или \(\cos x + \sin x = 2 + \sqrt{3}\)

или \(\cos x + \sin x = 2 - \sqrt{3}\)

\(\operatorname{tg} x = - 1\) или \(\sin\left( x + \frac{\pi}{4} \right) = \sqrt{2} + \sqrt{\frac{3}{2}}\) – корней нет;

или \(\sin\left( x + \frac{\pi}{4} \right) = \sqrt{2} - \sqrt{\frac{3}{2}}\)

\(x_{1} = - \dfrac{\pi}{4} + \pi n,\)
\(x_{2} = (-1)^{n}\arcsin\dfrac{2 - \sqrt{3}}{\sqrt{2}} - \dfrac{\pi}{4} + \pi n, n \in Z.\)