Question

已知函数f(x)=-2sin^2x+2根号3sinxcosx+1

Answer 1

sin^2 X = 1/2 - (1/2)cos(2X))

so f(x) = -1 + cos(2x) + sqrt(3)sin2x + 1

= 2sin(2x+pi/6)

when x belongs to -pi/6,pi/3,

2x + pi/6 belongs to -pi/6 to 5pi/6

fmin = -0.5

fmax = 1