2 корня из 2 sin^2 2,5x+sin3x=cos3x+корень из 2

[latex]2 \sqrt{2}\cdot ( \frac{1-cos5x}{2})+sin3x-cos3x- \sqrt{2}=0 \\ \\ \sqrt{2}- \sqrt{2}cos5x+ \sqrt{2}\cdot ( \frac{1}{ \sqrt{2} }sin3x- \frac{1}{ \sqrt{2} }cos3x)- \sqrt{2} =0 \\ \\ - \sqrt{2}cos5x+ \sqrt{2}\cdot ( sin \frac{ \pi }{ 4 }\cdot sin3x-cos \frac{ \pi }{ 4}\cdot cos3x)=0 [/latex] [latex] - \sqrt{2}cos5x- \sqrt{2}cos(3x+ \frac{ \pi }{ 4 })=0 \\ \\ cos5x+cos(3x+ \frac{ \pi }{ 4 })=0 \\ \\ 2cos \frac{5x+3x+ \frac{ \pi }{4} }{2}\cdot cos \frac{5x-3x- \frac{ \pi }{4} }{2}=0 \\ \\ cos (4x+ \frac{ \pi }{8} )\cdot cos (x- \frac{ \pi }{8} )=0 [/latex] cos(4x+(π/8))=0  ⇒  4x+(π/8) =(π/2)+πk, k∈Z ⇒   4x=(π/2)- (π/8) +πk, k∈Z ⇒ x=(3π/32) +(π/4)·k, k∈Z или  cos(x-(π/8))=0  ⇒  x =(π/2)+(π/8)+πn, n∈Z  ⇒   x=(5π/8) +(π/4)·n, n∈Z . О т в е т. x=(3π/32) +(π/4)·k ;   x=(5π/8) +(π/4)·n,  k, n∈Z .