已知sin^2α+√3sinαcosα-2cos^2a=0 α∈(π/6,5π/12)求sin(2α-π/3)和cos2α的值

sin2α + √3sinαcosα – 2cos2α = 0 => tan2α + √3tanα – 2 = 0 => Δ= 3 + 8 = 11 => tanα = (-√3 + √11)/2（因为α∈(π/6,5π/12),所以tanα > 0,负值舍去）所以,sin2α = 2tanα/(1 + tan2α) = (-√3 + √11)/[1 + (14 – 2√33)/4] = 2(-√3 + √11)/(9 – √33) = (√11 - √3)(9 + √33)/24 = (3√11 + √3)/12 ；cos2α = (1 – tan2α)/(1 + tan2α) = [1 - (14 – 2√33)/4]/[1 + (14 – 2√33)/4] = (√33 – 5)/(9 – √33) = (√33 – 5)(9 + √33)/48 = (9√33 – 45 + 33 - 5√33)/48 = (√33 – 3)/12；
而sin(2α - π/3) = (1/2)sin2α – (√3/2)cos2α = (3√11 + √3)/24 - (3√11 – 3√3)/24 = √3/6 .