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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 !