The minimum positive period of the function y = 2Sin (2x + π / 3) + sin (2x - π / 3)

The minimum positive period of the function y = 2Sin (2x + π / 3) + sin (2x - π / 3)

y=2sin(2x+π/3)+sin(2x-π/3)
=2 (sin2xcos / 3 + cos2xsin / 3) + sin2xcos / 3-cos2xsin / 3
=3 sin2xcos / 3 + cos2xsin / 3
=3 / 2sin2x + radical 3 / 2cos2x
From this we can see that the minimum positive period is Pai

Know the function f (x) = - 2Sin ^ 2 (3 / 4 + 2x) - root sign 3 cos4x, find the minimum positive period of 1, f (x), 2, the value of X when f (x) reaches the maximum value

(3 π / 4 + 2x) - 2 (3 π / 4 + 2x) - 2 (3 π / 4 + 2x) - 2 (3 π / 4 + 4 + 2x) - 2 (3 π / 4 + 2 + 2x) - √ 3cos4x = cos (3 π / 2 + 4x) - 1 - √ 3cos4x = sin4x = sin4x - √ 3cos4x-1 = 2 (1 / 2 * sin4x - √ 3 / 2 * cos4x-1) = 2 (sin4x cos π / 3-cos4x / 3) - 1 = 2Sin (4x - π / 3) - 1 = 2Sin (4x - π / 3) - 1 = 2Sin (4x - π / 3) - 1 = 2Sin (4x - π / 3) - 1 = 2 = 2 = 2 1t = 2 π / 4 = π / 2 when sin (4x - π / 3)

Function f (x)= 2+1 The value range of x2 − 2x + 3 is______ .

∵ let g (x) = x2-2x + 3 = (x-1) 2 + 2 ≥ 2,
∴g（x）min=g（1）=2，
∴f（x）max=
2+1
2=3
Two
2，
When G (x) → + ∞, f (x) →
2，
So the answer is:（
2，3
Two
2]．

Find the value range of the function f (x) = root (4-x) - root (2x + 1)

The definition domain is 4-x > = 0,2x + 1 > = 0, that is - 1 / 2 < = x < = 4
The function Y1 = radical (4-x) is a subtraction function, and y2 = - radical (2x + 1) is also a subtraction function
Therefore, f (x) = Y1 + Y2 is also a minus function
So the minimum value is f (4) = 0 - radical 9 = - 3
The maximum value is f (- 1 / 2) = root sign (4 + 1 / 2) - 0 = 3 / 2 root sign 2
Therefore, the value range is [- 3,3 / 2 radical sign 2]

Find the value range of 1-2x function under y = x + quadratic root sign

Let a = √ (1-2x)
Then a > = 0
a²=1-2x
x=(1-a²)/2
So y = (1-A 2) / 2 + a
=-1/2a²+a+1/2
=-1/2(a-1)²+1
a>=0
So a = 1, y max = 1
So the range of values is (- ∞, 1]

What is the minimum positive period of the function y = cos ^ 2x sin ^ 2x + 2sinxcosx

y=cos^2x-sin^2x+2sinxcosx
=(2cos^2x-1)+sin2x
=cos2x+sin2x
=√2sin(2x+∏/4）
Minimum positive period Π
The range is [- √ 2, √ 2]

The value range of function y = 1 / 2Sin) 2x) + sin ^ 2 (x), X ∈ R

Because cos (2x) = 1-2 sin ^ 2 (x),
So sin ^ 2 (x) = [1-cos (2x)] / 2
y=1/2sin(2x)+sin^2(x)
=1/2sin(2x)+ [1-cos(2x)]/2
=1/2*sin(2x)- 1/2*cos(2x)+1/2
=√2/2*[√2/2* sin(2x)- √2/2* cos(2x)] +1/2
=√2/2* sin(2x-π/4) +1/2
The value range of the function is [- √ 2 / 2 + 1 / 2, √ 2 / 2 + 1 / 2]

Given the function f (x) = cos (X-2 π / 3) - cosx, find the minimum positive period and monotone increasing interval of the function

cos(x-2π/3)-cosx
=cosxcos(2π/3)+sinxsin(2π/3)-cosx
=-(3/2)cosx+(√3/2)sinx
=√3[sinx*cos(π/3)-cosx*sin(π/3)]
=√3sin(x-π/3)
（1）T =2π
（2） 2kπ-π/2 ≤x-π/3≤2kπ+π/2
2kπ-π/6 ≤x≤2kπ+5π/6
So we increase the interval [2K π - π / 6, 2K π + 5 π / 6], K ∈ Z

Simplify f (x) = cos ((6K + 1) / 3 * π + 2x) + cos ((6k-1) / 3 * π - 2x) (x ∈ R, K ∈ z), and find the range and minimum positive period of function f (x)

(6K + 1) π / 3 + 2x] = cos [2K + 1) π / 3 + 2x] = cos [2K π + π / 3 + 2x] = cos [π / 3 + 2x] cos [(6k-1) π / 3-2x] = cos [π [2K π - π / 3-2x] = cos [π / 3 + 2x] so the formula = 2cos [π / 3 + 2x] + 2 √ 3sin (π / 6-2x) 2 √ 3sin (π / 6-2x) 2 √ 3sin (π / 6-2x) = 2 √ 3sin [π [π [6-6-2x) = 2 √ 3sin [π [π [π [6 2 - (π / 3 + 2x)] = 2 √ 3cos (π / 3 + 2x)

Let f (x) = cos ω x (√ 3sin ω x + cos ω x), where 0

Because f (x) = cos ω x (√ 3sin ω x + cos ω x) = √ 3sin ω xcoswx + (coswx) ^ 2 = √ 3 / 2 × sin2wx + 1 / 2 × cos2wx + 1 / 2
=Sin (2wx + π / 6) + 1 / 2 and 0