Question

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

Answer 1

$2 sqrt{2}cdot ( frac{1-cos5x}{2})+sin3x-cos3x- sqrt{2}=0 \ \ sqrt{2}- sqrt{2}cos5x+ sqrt{2}cdot ( frac{1}{ sqrt{2} }sin3x- frac{1}{ sqrt{2} }cos3x)- sqrt{2} =0 \ \ - sqrt{2}cos5x+ sqrt{2}cdot ( sin frac{ pi }{ 4 }cdot sin3x-cos frac{ pi }{ 4}cdot cos3x)=0$

$- sqrt{2}cos5x- sqrt{2}cos(3x+ frac{ pi }{ 4 })=0 \ \ cos5x+cos(3x+ frac{ pi }{ 4 })=0 \ \ 2cos frac{5x+3x+ frac{ pi }{4} }{2}cdot cos frac{5x-3x- frac{ pi }{4} }{2}=0 \ \ cos (4x+ frac{ pi }{8} )cdot cos (x- frac{ pi }{8} )=0$

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 .

Answer 2

cosx=0. x= п/2 или х= -п/2