Упр.673 ГДЗ Алимов 10-11 класс (Алгебра)

2) sin2 x + cos2 2x = 1;
3) sin 4x = 6 cos2 2x — 4;
4) 2 cos2 3x + sin 5x — 1

1. $$\sin^2 x+\sin^2 2x=1$$

   $$\sin^2 2x=1-\sin^2 x=\cos^2 x$$

   $$4\sin^2 x\cos^2 x=\cos^2 x$$

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

   Отсюда:

   $$\cos x=0 \quad \text{или} \quad 4\sin^2 x-1=0$$

   1) $$\cos x=0 \Rightarrow x=\frac{\pi}{2}+\pi n,\; n\in\mathbb Z$$

   2) $$4\sin^2 x-1=0 \Rightarrow \sin x=\pm \frac12$$

   $$x=\pm \frac{\pi}{6}+\pi n,\; n\in\mathbb Z$$

2. $$\sin^2 x+\cos^2 2x=1$$

   $$\cos^2 2x=1-\sin^2 x=\cos^2 x$$

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

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

   $$4\cos^4 x-5\cos^2 x+1=0$$

   Пусть $$y=\cos^2 x$$. Тогда:

   $$4y^2-5y+1=0$$

   $$D=25-16=9$$

   $$y_1=\frac{5-3}{8}=\frac14,\qquad y_2=\frac{5+3}{8}=1$$

   1) $$\cos^2 x=1 \Rightarrow \cos x=\pm 1 \Rightarrow x=\pi n,\; n\in\mathbb Z$$

   2) $$\cos^2 x=\frac14 \Rightarrow \cos x=\pm \frac12 \Rightarrow x=\pm \frac{\pi}{3}+\pi n,\; n\in\mathbb Z$$

3. $$\sin 4x=6\cos^2 2x-4$$

   $$2\sin 2x\cos 2x=6\cos^2 2x-4(\cos^2 2x+\sin^2 2x)$$

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

   Разделим на $$\cos^2 2x$$:

   $$2\tan 2x-2+4\tan^2 2x=0$$

   Пусть $$y=\tan 2x$$. Тогда:

   $$4y^2+2y-2=0$$

   $$2y^2+y-1=0$$

   $$D=1+8=9$$

   $$y_1=-1,\qquad y_2=\frac12$$

   1) $$\tan 2x=-1 \Rightarrow 2x=-\frac{\pi}{4}+\pi n \Rightarrow x=-\frac{\pi}{8}+\frac{\pi n}{2}$$

   2) $$\tan 2x=\frac12 \Rightarrow 2x=\arctan \frac12+\pi n \Rightarrow x=\frac12\arctan \frac12+\frac{\pi n}{2}$$

4. $$2\cos^2 3x+\sin 5x=1$$

   $$2\cos^2 3x+\sin 5x-1=0$$

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

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

   $$\cos 6x+\cos\left(\frac{\pi}{2}-5x\right)=0$$

   $$2\cos\left(3x+\frac{\pi}{4}-\frac{5x}{2}\right)\cos\left(3x-\frac{\pi}{4}+\frac{5x}{2}\right)=0$$

   $$\cos\left(\frac{x}{2}+\frac{\pi}{4}\right)\cos\left(\frac{11x}{2}-\frac{\pi}{4}\right)=0$$

   Отсюда:

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

   или

   $$\cos\left(\frac{11x}{2}-\frac{\pi}{4}\right)=0 \Rightarrow \frac{11x}{2}-\frac{\pi}{4}=\frac{\pi}{2}+\pi n \Rightarrow x=\frac{3\pi}{22}+\frac{2\pi n}{11}$$

Ответ

1) $$x=\frac{\pi}{2}+\pi n,\; x=\pm \frac{\pi}{6}+\pi n,\; n\in\mathbb Z$$

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

3) $$x=-\frac{\pi}{8}+\frac{\pi n}{2},\; x=\frac12\arctan \frac12+\frac{\pi n}{2},\; n\in\mathbb Z$$

4) $$x=\frac{\pi}{2}+2\pi n,\; x=\frac{3\pi}{22}+\frac{2\pi n}{11},\; n\in\mathbb Z$$