Question

Sin^2*x + sin^2*4x=sin^2*2x + sin^2*3x

Answer 1

(1-cos2x)/2+(1-cos8x)/2=(1-cos4x)/2+(1-cos6x)/2
1-cos2x+1-cos8x+cos4x+cos6x-1-1=0
(cos6x-cos8x)+(cos2x-cos4x)=0
2sin7xsinx+2sin3xsinx=0
2sinx(sin7x+sin3x)=0
4sinxsin5xcos2x=0
sinx=0⇒x=πn,n∈z
sin5x=0⇒5x=πk,k∈z⇒x=πk/5,k∈z
cos25x=0⇒2x=π/2+πm,m∈z⇒x=π/4+πm/2,m∈z
ответ x=πk/5,k∈z;x=π/4+πm/2,m∈z