Упр.22.23 ГДЗ Мордкович 10-11 класс (Алгебра)

1. Решите уравнение: а) $$\sin 3x=\cos 2x$$; б) $$\sin(5\pi-x)=\cos(2x+7\pi)$$; в) $$\cos 5x=\sin 15x$$; г) $$\sin(7\pi+x)=\cos(9\pi+2x)$$.

а) $$\sin 3x=\cos 2x$$

$$\sin 3x-\cos 2x=0$$

$$\sin 3x-\sin\left(\frac{\pi}{2}-2x\right)=0$$

$$2\cos\frac{3x+\frac{\pi}{2}-2x}{2}\cdot \sin\frac{3x-\left(\frac{\pi}{2}-2x\right)}{2}=0$$

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

Отсюда

$$\cos\left(\frac{x}{2}+\frac{\pi}{4}\right)=0 \quad \text{или} \quad \sin\left(\frac{5x}{2}-\frac{\pi}{4}\right)=0$$

1) $$\frac{x}{2}+\frac{\pi}{4}=\frac{\pi}{2}+\pi n \Rightarrow x=\frac{\pi}{2}+2\pi n$$

2) $$\frac{5x}{2}-\frac{\pi}{4}=\pi n \Rightarrow x=\frac{\pi}{10}+\frac{2\pi n}{5}$$

Ответ: $$x=\frac{\pi}{2}+2\pi n,\; x=\frac{\pi}{10}+\frac{2\pi n}{5},\; n\in\mathbb Z$$

б) $$\sin(5\pi-x)=\cos(2x+7\pi)$$

$$\sin(\pi-x)=\cos(2x+\pi)$$

$$\sin x=-\cos 2x$$

$$\sin x+\cos 2x=0$$

$$\cos\left(\frac{\pi}{2}-x\right)+\cos 2x=0$$

$$2\cos\left(\frac{\frac{\pi}{2}-x+2x}{2}\right)\cos\left(\frac{\frac{\pi}{2}-x-2x}{2}\right)=0$$

$$2\cos\left(\frac{x}{2}+\frac{\pi}{4}\right)\cos\left(\frac{3x}{2}-\frac{\pi}{4}\right)=0$$

Отсюда

$$\cos\left(\frac{x}{2}+\frac{\pi}{4}\right)=0 \quad \text{или} \quad \cos\left(\frac{3x}{2}-\frac{\pi}{4}\right)=0$$

1) $$\frac{x}{2}+\frac{\pi}{4}=\frac{\pi}{2}+\pi n \Rightarrow x=\frac{\pi}{2}+2\pi n$$

2) $$\frac{3x}{2}-\frac{\pi}{4}=\frac{\pi}{2}+\pi n \Rightarrow x=\frac{\pi}{2}+\frac{2\pi n}{3}$$

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

в) $$\cos 5x=\sin 15x$$

$$\cos 5x-\sin 15x=0$$

$$\cos 5x-\cos\left(\frac{\pi}{2}-15x\right)=0$$

$$-2\sin\left(\frac{5x+\frac{\pi}{2}-15x}{2}\right)\sin\left(\frac{5x-\left(\frac{\pi}{2}-15x\right)}{2}\right)=0$$

$$2\sin\left(5x-\frac{\pi}{4}\right)\sin\left(10x-\frac{\pi}{4}\right)=0$$

Отсюда

$$\sin\left(5x-\frac{\pi}{4}\right)=0 \quad \text{или} \quad \sin\left(10x-\frac{\pi}{4}\right)=0$$

1) $$5x-\frac{\pi}{4}=\pi n \Rightarrow x=\frac{\pi}{20}+\frac{\pi n}{5}$$

2) $$10x-\frac{\pi}{4}=\pi n \Rightarrow x=\frac{\pi}{40}+\frac{\pi n}{10}$$

Ответ: $$x=\frac{\pi}{20}+\frac{\pi n}{5},\; x=\frac{\pi}{40}+\frac{\pi n}{10},\; n\in\mathbb Z$$

г) $$\sin(7\pi+x)=\cos(9\pi+2x)$$

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

$$-\sin x=-\cos 2x$$

$$\cos 2x-\sin x=0$$

$$\cos 2x-\cos\left(\frac{\pi}{2}-x\right)=0$$

$$-2\sin\left(\frac{2x+\frac{\pi}{2}-x}{2}\right)\sin\left(\frac{2x-\left(\frac{\pi}{2}-x\right)}{2}\right)=0$$

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

Отсюда

$$\sin\left(\frac{x}{2}+\frac{\pi}{4}\right)=0 \quad \text{или} \quad \sin\left(\frac{3x}{2}-\frac{\pi}{4}\right)=0$$

1) $$\frac{x}{2}+\frac{\pi}{4}=\pi n \Rightarrow x=-\frac{\pi}{2}+2\pi n$$

2) $$\frac{3x}{2}-\frac{\pi}{4}=\pi n \Rightarrow x=\frac{\pi}{6}+\frac{2\pi n}{3}$$

Ответ: $$x=-\frac{\pi}{2}+2\pi n,\; x=\frac{\pi}{6}+\frac{2\pi n}{3},\; n\in\mathbb Z$$