Question

Решите уравнения sin2x=1 cosx*cos2x+sinx*sin2x=0 cos^2x=cos2x

Answer 1

1.

$\sin2x=1$

$2x=\dfrac{\pi}{2}+2\pi n$

$x=\dfrac{\pi}{4}+\pi n ,\ n\in\mathbb{Z}$

2.

$\cos x\cdot \cos2x+\sin x\cdot\sin2x=0$

$\cos( x-2x)=0$

$\cos( -x)=0$

$\cos x=0$

$x=\dfrac{\pi}{2}+\pi n,\ n\in\mathbb{Z}$

3.

$\cos^2x=\cos2x$

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

$-\sin^2x=0$

$\sin^2x=0$

$\sin x=0$

$x=\pi n,\ n\in\mathbb{Z}$

Answer 2

https://znanija.com/task/37922892

Решите уравнения :

1. sin2x =1  ⇒  2x = π/2 +2πn , n∈ ℤ    ⇔ x  = π/4 +πn  , n∈ ℤ .

2.cos2x*cosx +sin2x*sinx =0⇔cos(2x-x) =0 ⇔cosx =0⇒ x  = π/2 +πn , n∈ ℤ .

3. cos²x =cos2x ⇔(1+cos2x)/2 =cos2x ⇔cos2x=1. ⇒ 2x  =2πn , n∈ ℤ  ;

x=πn  ,  n∈ ℤ .