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

а) sin 3x = cos 2x;
б) sin (5пи — x) = cos (2x + 7пи);
в) cos 5x = sin 15x;
г) sin (7пи + x) = cos (9пи + 2x).

1. $$\sin 3x=\cos 2x$$

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

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

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

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

   Отсюда:

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

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

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

2. $$\sin(5\pi-x)=\cos(2x+7\pi)$$

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

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

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

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

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

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

   Отсюда:

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

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

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

3. $$\cos 5x=\sin 15x$$

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

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

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

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

   Отсюда:

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

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

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

4. $$\sin(7\pi+x)=\cos(9\pi+2x)$$

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

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

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

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

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

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

   Отсюда:

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

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

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

Ответ

а) $$x=\frac{\pi}{2}+2\pi n$$ или $$x=\frac{\pi}{10}+\frac{2\pi n}{5}$$;

б) $$x=\frac{\pi}{2}+2\pi n$$ или $$x=\frac{\pi}{2}+\frac{2\pi n}{3}$$;

в) $$x=\frac{\pi}{40}+\frac{\pi n}{10}$$ или $$x=\frac{\pi}{20}+\frac{\pi n}{5}$$;

г) $$x=-\frac{\pi}{2}+2\pi n$$ или $$x=\frac{\pi}{6}+\frac{2\pi n}{3}$$, $$n\in\mathbb{Z}$$.