2sinx = 2cosx + √6
2•(sinx - cosx) = √6
2•√2•( (√2/2)•sinx - (√2/2)•cosx) ) = √6
2•√2•( cos(π/4)sinx - sin(π/4)cosx ) = √6
sinα•cosβ - cosα•sinβ = sin(α - β)
2√2•sin(x - (π/4)) = √6
sin(x - (π/4)) = √3/2
[ x - (π/4) = (π/3) + 2πn ⇔ x = (7π/12) + 2πn , n ∈ Z
[ x - (π/4) = (2π/3) + 2πk ⇔ x = (11π/12) + 2πk , k ∈ Z
ОТВЕТ: (7π/12) + 2πn , n ∈ Z ; (11π/12) + 2πk , k ∈ Z