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.
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|2sin5x +sin9x+sinx | ≤|2sin5x| +|sin9x|+|sinx| ≤ 2*1+1+1 = 4