4sin3x·sin5x·sin8x=sin6x ; x∈[-π/2;7π/2] ⇔ x∈[-π/2;3π+π/2]
4sin3x·sin5x·s-n8x - 2sin3x·cos3x = 0
sin3x(2sin5x·sin8x - cos3x) = 0
sin3x=0 ⇒ 3x = πk ⇒ x = π/3·k ⇒
⇒ x={0; π/3;2π/3;π;4π/3;5π/3;2π;7π/3}
2sin5x·sin8x - cos3x =0
sinα·sinβ = 1/2[cos(α-β) - cos(α+β)] ⇒
(cos3x - cos13x) -cos3x =0 ⇒
cos13x=0 ⇒ 13x = +/-π/2 +2πn ; n ∈Z
x= +/- π/26 + 2πn/13 ; n∈ Z
a) -π/ 2 ≤ -π/26 +2πn/13≤7π/2
b) -π/2 ≤ π/26 +2πn/13≤7π/2
n находите сами !