Страница 411 номер 1367, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

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

2) \(\sin{4x} = \cos^{4}x - \sin^{4}x\)
\(2\sin{2x}\cos{2x} - \left( \cos^{2}x - \sin^{2}x \right)\left( \cos^{2}x + \sin^{2}x \right) = 0,\)
\(\cos{2x}\left( 2\sin{2x} - 1 \right) = 0,\)
\(\cos{2x} = 0\) или \(\sin{2x} = \frac{1}{2}\)
\(x_{1} = \dfrac{\pi}{4} + \dfrac{\pi}{2}n, n \in Z\)
\(x_{2} = (-1)^{n} \cdot \frac{\pi}{12} + \frac{\pi}{2}n\), \(n \in Z\);

3) \(2\cos^{2}x = 1 + 4\sin{2x}\)
\(2\cos^{2}x - 1 = 4\sin{2x},\)
\(\cos{2x} = 4\sin{2x},\)
\(\operatorname{ctg} {2x} = 4,\)
\(\operatorname{tg} {2x} = \dfrac{1}{4},\)
\(2x = \operatorname{arctg} \dfrac{1}{4} + \pi n,\)
\(x = \dfrac{1}{2}\operatorname{arctg} \dfrac{1}{4} + \dfrac{\pi}{2}n, n \in Z;\)

4) \(2\cos x + \cos{2x} = 2\sin x\)
\(2\left( \cos x - \sin x \right) + \left( \cos^{2}x - \sin^{2}x \right) = 0,\)
\(\left( \cos x - \sin x \right)\left( 2 + \cos x + \sin x \right) = 0,\)
\(\cos x - \sin x = 0\) или \(\cos x + \sin x = - 2\) – корней нет;
\(\cos x + \sin x = \sqrt{2}\cos\left( x - \dfrac{\pi}{4} \right) \geqslant - \sqrt{2} > - 2;\)
\(\operatorname{tg} x = 1,\)
\(x = \dfrac{\pi}{4} + \pi n, n \in Z.\)