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

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