Question

cos²2x+cos²6x-1=0 хелппппп

Answer 1

сos^2(2x)+cos^2(6x)-1=0
(1+cos4x)/2  + (1+cos12x)/2 = 1
1+cos4x + 1 + cos12x = 2
cos4x+cos12x=0
2cos((4x+12x)/2)cos((12x-4x)/2)=0
2cos8xcos4x=0

cos8x=0 ;              cos4x=0
8x= п/2 + пk         4x= п/2 + пk
x = п/16 + пk/8     x= п/8 + пk/4

 

Answer 2

((1+cosx)/2+(1+cos3x)/2)-1=0;
1+cosx+1+cos3x-2=0;
cos3x+cosx=0;
2cos(3x+x)/2*cos(3x-x)/2=0;
2cos*2xcosx=0;
cos2x=0 (1); cosx=0 (2)

(1) 2x= π/2+πn; x=π/4+πn/2
(2) x=π/2+πn

 

Ответ: x= π/4+πn/2; x=π/2+πn