Sin5x * cos^2 2x = 1

sin5x *cos²2x =1 ; sin5x *(1+cos2*2x) /2 = 1 ; sin5x +sin5x*cos4x =2 ; sin5x +( sin(5x +4x) +sin(5x-4x) ) / 2 = 2 ; 2sin5x +sin9x +sinx = 4 ⇔ { sinx =1 ; sin5x = 1; sin9x =1.⇔  { x =π/2 +2πk ; 5x = π/2 +2πm ; 9x =π/2 +2πn ,k,m,n ∈ Z. ⇒ x =π/2 +2πk ; x = (π/2 +2πm)/5 ; x =(π/2 +2πn)/9 ,k,m,n ∈ Z. ⇔x =π/2 +2πk ,  k ∈ Z. * * * (π/2 +2πm)/5=π/2 +2πk  ⇔2πm =2π +10πk ⇔m=1+5k * * * * * * (π/2 +2πn)/9 =π/2 +2πk  ⇔2πn = 4π +18πk ⇔n=2+9k * * * ответ : π/2 +2πk ,  k ∈ Z.  * * * * * * * cos² α/2 =(1+cos2α)/2 ; sinα*cosβ  =( sin(α+β) +  sin(α-β) ) /2.    --- |2sin5x +sin9x+sinx | ≤|2sin5x| +|sin9x|+|sinx|  ≤  2*1+1+1 = 4