2sinx=2cosx+корень из 6

2sinx = 2cosx + √62•(sinx - cosx) = √62•√2•( (√2/2)•sinx - (√2/2)•cosx) ) = √62•√2•( cos(π/4)sinx - sin(π/4)cosx ) = √6

sinα•cosβ - cosα•sinβ = sin(α - β)

2√2•sin(x - (π/4)) = √6sin(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 ∈ ZОТВЕТ: (7π/12) + 2πn , n ∈ Z ; (11π/12) + 2πk , k ∈ Z