Упр.16 Повторение ГДЗ Мордковича 11 класс профильный уровень (Алгебра)

а) $$\cos 2x+(\sin x+\cos x)^2\cdot \tg x=\tg x\cdot(\tg x+1);$$

$$\cos 2x+(\sin^2 x+2\sin x\cos x+\cos^2 x)\cdot \tg x=\tg x\cdot(\tg x+1);$$

$$\left(1-2\sin^2 x\right)+\left(1+2\sin x\cos x\right)\cdot \frac{\sin x}{\cos x}=\tg x\cdot(\tg x+1);$$

$$1-2\sin^2 x+\frac{\sin x}{\cos x}+2\sin^2 x=\tg x\cdot(\tg x+1);$$

$$1+\tg x=\tg^2 x+\tg x;$$

$$\tg^2 x=1;$$

$$\tg x=\pm 1;$$

$$x=\pm \frac{\pi}{4}+\pi n,\quad n\in \mathbb{Z}.$$

На отрезке $$\left[-\frac{7\pi}{4};\frac{\pi}{4}\right]$$ получаем:

$$x=-\frac{7\pi}{4},\ -\frac{5\pi}{4},\ -\frac{3\pi}{4},\ -\frac{\pi}{4},\ \frac{\pi}{4}.$$

Ответ: $$-\frac{7\pi}{4};\ -\frac{5\pi}{4};\ -\frac{3\pi}{4};\ -\frac{\pi}{4};\ \frac{\pi}{4}.$$

б) $$\sin\left(\frac{\pi}{4}-4x\right)\cos\left(\frac{\pi}{4}-x\right)+\sin^2\frac{5x}{2}=0;$$

$$\frac12\left(\sin\left(\frac{2\pi}{4}-5x\right)+\sin(-3x)\right)+\sin^2\frac{5x}{2}=0;$$

$$\frac12\sin\left(\frac{\pi}{2}-5x\right)-\frac12\sin 3x+\frac{1-\cos 5x}{2}=0;$$

$$\frac12\cos 5x-\frac12\sin 3x+\frac12-\frac12\cos 5x=0;$$

$$\frac12\sin 3x=\frac12;$$

$$\sin 3x=1;$$

$$3x=\frac{\pi}{2}+2\pi n,\quad n\in \mathbb{Z};$$

$$x=\frac{\pi}{6}+\frac{2\pi n}{3},\quad n\in \mathbb{Z}.$$

На отрезке $$[-\pi;\pi]$$:

$$x=-\frac{\pi}{2},\ \frac{\pi}{6},\ \frac{5\pi}{6}.$$

Ответ: $$-\frac{\pi}{2};\ \frac{\pi}{6};\ \frac{5\pi}{6}.$$