Страница 192 номер 631, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(2\sin{2x} - 3\left( \sin x + \cos x \right) + 2 = 0\)
\(2\sin{2x} - 3\left( \sin x + \cos x \right) + 2\left( \sin^{2}x + \cos^{2}x \right) = 0,\)
\(2\sin{2x} - 3\left( \sin x + \cos x \right) + 2\left( \sin x + \cos x \right)^{2} - 4\sin x\cos x = 0,\)
\(\left( \sin x + \cos x \right)\left( 2\sin x + 2\cos x - 3 \right) = 0,\)
\(\operatorname{tg} x + 1 = 0\) или \(\sin x + \cos x = \frac{3}{2}\)
\(\operatorname{tg} x = - 1\) или \(\frac{\sqrt{2}}{2}\sin x + \frac{\sqrt{2}}{2}\cos x = \frac{3}{2\sqrt{2}}\) – корней нет,

т.к. \(\sin\left( x + \frac{\pi}{4} \right) = \frac{3}{2\sqrt{2}} > 1\)

\(x = - \dfrac{\pi}{4} + \pi n, n \in Z;\)

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

3) \(\sin{2x} + 4\left( \sin x + \cos x \right) + 4 = 0\)
\(2\sin x\cos x + 4\left( \sin x + \cos x \right) + 3 + \sin^{2}x + \cos^{2}x = 0,\)
\(4\left( \sin x + \cos x \right) + 3 + \left( \sin x + \cos x \right)^{2} = 0,\)
\(\sin x + \cos x = z, z^{2} + 4z + 3 = 0,\)
\(z_{1{,}2} = \dfrac{-4 \pm \sqrt{16 - 12}}{2},\)
\(z_{1} = - 1, z_{2} = - 3\)
\(\sin x + \cos x = - 1\) или \(\sin x + \cos x = - 3\)
\(\frac{\sqrt{2}}{2}\sin x + \frac{\sqrt{2}}{2}\cos x = - \frac{\sqrt{2}}{2}\) или \(\frac{\sqrt{2}}{2}\sin x + \frac{\sqrt{2}}{2}\cos x = - \frac{3}{\sqrt{2}}\)
\(\sin\left( x + \frac{\pi}{4} \right) = - \frac{\sqrt{2}}{2}\) или \(\sin\left( x + \frac{\pi}{4} \right) = - \frac{3}{\sqrt{2}}\) – корней нет;
\(x + \dfrac{\pi}{4} = (-1)^{n + 1} \cdot \dfrac{\pi}{4} + \pi n,\)
\(x = (-1)^{n + 1} \cdot \dfrac{\pi}{4} - \dfrac{\pi}{4} + \pi n, n \in Z;\)

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