task/29523893
366. 2) sin2x*cosx = cos2x*sinx ⇔sin2x*cosx - cos2x*sinx=0⇔sin(2x -x) =0 ⇔
sinx = 0⇒ x =πn , n∈ℤ .
366. 4) cos5x*cosx =cos4x⇔[cos(5x-x) +cos(5x+x)] / 2 =cos4x⇔
cos4x +cos6x=2cos4x⇔cos6x- cos4x = 0 ⇔ -2sin[(6x+4x)/2]sin[(6x-4x)/2] =0 ⇔
[ sin5x =0 ; sinx =0 .⇔[ 5x =πn ; x= πn , n∈ℤ .⇔[ x =(π/5)*n ; x= πn , n∈ℤ .
Удачи !