Question

Решите уравнение , Пожалуйста
1.sin^2 x+sin^2 3x=1
2. sin 3x = cos 2x
3. sin ( 5pi - x) = сos (2x + 7pi)

Answer 1

1
(1-cos2x)/2+(1-cos6x)/2=1
1-cos2x+1-cos6x=2
cos2x+cos6x=0
2cos4ccos2x=0
cos4x=0⇒4x=π/2+πn,n∈z⇒x=π/8+πn/4,n∈z
cos2x=0⇒2x=π/2+πk,k∈z⇒x=π/4+πk/2,k∈z
2
sin3x-sin(π/2-2x)=0
2sin(5x/2-π/4)cos(x/2+π/4)=0
sin(5x/2-π/4)=0⇒5x/2-π/4=πn,n∈z⇒5x/2=π/4+πn,n∈z⇒
x=π/10+2πn/5,n∈z
cos(x/2+π/4)=0⇒x/2+π/4=π/2+πk,k∈x⇒x/2=π/4+πk,k∈z⇒
x=π/2+2πk,k∈z
3
cosx+cos2x=0
2xos3x/2cosx/2=0
cos3x/2=0⇒3x/2=π/2+πn,n∈z⇒x=π/3+2πn/3,n∈z
cosx/2=0⇒x/2=π/2+πk,k∈z⇒x=π|2πk,k∈z