Given the function f (x) = 2sinxcosx + 2cosx & # 178; (x ∈ R) 1, find the value 2 of F (x) and the range of F (x) when x ∈ [0, π / 2]

Given the function f (x) = 2sinxcosx + 2cosx & # 178; (x ∈ R) 1, find the value 2 of F (x) and the range of F (x) when x ∈ [0, π / 2]

(1)f(x)=sin2x+cos2x+1
=√2(sin2x *√/2+cos2x *√2)+1
=√2(sin2x cosπ/4+cos2x cosπ/2)+1
=√2sin(2x+π/4)+1
(2) When x ∈ [0, π / 2]
2x+π/4∈[π/4,3π/4]
sin(2x+π/4)∈[√2/2,1]
So y ∈ [2, √ 2 + 1]

Let f (x) = 2sinxcosx + 2cosx Square-1
1. Find the period and maximum of F (x)
2. Find the monotone increasing interval of F (x)

F (x) = the square of 2sinxcosx + 2cosx - 1 = sin2x + cos2x = √ 2Sin (2x + π / 4)
The period of F (x) = π
The maximum value of F (x) = 2
Monotone increasing interval [- 3 π / 8 + K π, π / 8 + K π]

Let f (x) = 2sinxcosx + 2cosx Square-1. Find the minimum positive period of F (x)

F (x) = the square of 2sinxcosx + 2cosx-1 = sin (2x) + cos (2x) = radical 2 [radical 2 / 2 · sin (2x) + radical 2 / 2 · cos (2x)] = radical 2 [cos (π / 4) · sin (2x) + radical 2 / 2 · cos (2x)] = radical 2Sin (2x + π / 4) = radical 2Sin (2 π + 2x + π / 4) = radical 2Sin [2 (π + x) + π / 4]