Given the function f (x) = 2sinxcosx-2cos ^ 2x, find the minimum positive period of the function. When x belongs to o, the value range of F (x) is 0,

f(x)=2sinxcosx-2cos²x
=sin2x-1-cos2x
=√2(√2/2 sin2x-√2/2 cos2x)-1
=√2(cosπ/4 sin2x-sinπ/4 cos2x)-1
=√2 sin(2x-π/4)-1
So the minimum positive period T = 2 π / 2 = π
When x ∈ [0, π / 2]
0≤2x≤π
-π/4≤2x-π/4≤3π/4
So when 2x - π / 4 = - π / 4, f (x) gets the minimum - √ 2 · √ 2 / 2-1 = - 2
When 2x - π / 4 = π / 2, the maximum value of F (x) is √ 2-1
So f (x) value range: [- 2, √ 2-1]

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