Упр.1382 ГДЗ Алимов 10-11 класс (Алгебра)

1. $$5+\sin 2x=5(\sin x+\cos x)$$
2. $$2+2\cos x=3\sin x\cos x+2\sin x$$

1) $$5+\sin 2x=5(\sin x+\cos x)$$

$$5+2\sin x\cos x=5(\sin x+\cos x)$$

$$4+(\sin x+\cos x)^2-5(\sin x+\cos x)=0$$

Пусть $$t=\sin x+\cos x$$. Тогда

$$t^2-5t+4=0$$

$$D=25-16=9$$

$$t_{1,2}=\frac{5\pm 3}{2}$$

$$t_1=4,\quad t_2=1$$

Если $$\sin x+\cos x=4$$, то решений нет, так как $$|\sin x+\cos x|\le \sqrt{2}$$.

Если $$\sin x+\cos x=1$$, то

$$\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)=1$$

$$\sin\left(x+\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$$

$$x+\frac{\pi}{4}=\frac{\pi}{4}+2\pi n \quad \text{или} \quad x+\frac{\pi}{4}=\frac{3\pi}{4}+2\pi n$$

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

2) $$2+2\cos x=3\sin x\cos x+2\sin x$$

$$1+\frac{3}{2}(\cos^2 x+\sin^2 x-2\sin x\cos x)+2(\cos x-\sin x)=0$$

$$3(\cos x-\sin x)^2+4(\cos x-\sin x)+1=0$$

Пусть $$t=\cos x-\sin x$$. Тогда

$$3t^2+4t+1=0$$

$$D=16-12=4$$

$$t_{1,2}=\frac{-4\pm 2}{6}$$

$$t_1=-1,\quad t_2=-\frac13$$

Если $$\cos x-\sin x=-1$$, то

$$\sqrt{2}\cos\left(x+\frac{\pi}{4}\right)=-1$$

$$\cos\left(x+\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}}$$

$$x+\frac{\pi}{4}=\frac{3\pi}{4}+2\pi n \quad \text{или} \quad x+\frac{\pi}{4}=\frac{5\pi}{4}+2\pi n$$

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

Если $$\cos x-\sin x=-\frac13$$, то

$$\sqrt{2}\cos\left(x+\frac{\pi}{4}\right)=-\frac13$$

$$\cos\left(x+\frac{\pi}{4}\right)=-\frac{1}{3\sqrt{2}}$$

$$x+\frac{\pi}{4}=\pm \arccos\left(-\frac{1}{3\sqrt{2}}\right)+2\pi n$$

$$x=-\frac{\pi}{4}\pm \arccos\left(-\frac{1}{3\sqrt{2}}\right)+2\pi n,\quad n\in\mathbb Z$$

Ответ: $$x=2\pi n,\ \frac{\pi}{2}+2\pi n;\ \frac{\pi}{2}+2\pi n,\ \pi+2\pi n,\ -\frac{\pi}{4}\pm \arccos\left(-\frac{1}{3\sqrt{2}}\right)+2\pi n,\ n\in\mathbb Z.$$