sin6x=2cos(3pi\2+2x)


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sin6x = 2cos((3π/2) + 2x)sin6x = 2sin2xsin6x = sin2x + sin2xsin6x - sin2x = sin2x

sinα - sinβ = 2•sin( (α-β)/2 )•cos( (α+β)/2 )

2•sin2x•cos4x = sin2xsin2x•(2cos4x - 1) = 01) sin2x = 0 ⇔ 2x = πn ⇔ х = πn/2 , n ∈ Z2) 2cos4x - 1 = 0 ⇔ cos4x = 1/2 ⇔ 4x = ± (π/3) + 2πk ⇔ x = ± (π/12) + (πk/2), k ∈ ZОТВЕТ: πn/2, n ∈ Z ; ± (π/12) + (πk/2), k ∈ Z