Let f (x) = 2cosx square + 2 radical sign 3sinxcosx-1 (1) find the period and monotone increasing interval of F (x) (2) It shows how the image of F (x) can be obtained by changing the image of y = SiNx

Let f (x) = 2cosx square + 2 radical sign 3sinxcosx-1 (1) find the period and monotone increasing interval of F (x) (2) It shows how the image of F (x) can be obtained by changing the image of y = SiNx

（1）
f(x)=2cos²x+2√3sinxcosx-1
=1+cos2x+√3sin2x-1
=cos2x+√3sin2x
=2（1/2cos2x+√3/2sin2x）
=2（sinπ/6cos2x+cosπ/6sin2x）
=2sin（π/6+2x）
Period T = 2 π / ω = 2 π / 2 = π
X ∈ [K π - π / 3, K π + π / 6], monotonically increasing
(2) The image of F (x) can be obtained from the image of y = SiNx, where x is doubled, y is doubled, and then π / 6 is shifted to the left

Let f (x) = 2cosx ^ 2 + 2 radical sign 3sinxcosx Find the maximum and minimum period of F (x) To process!

f(x）=2cos^2x+2√3sinxcosx
=1+cos2x+√3sin2x
=1+2sin(π/6+2x)
So the maximum is 3
The minimum value is - 1
The period is 2 π / 2 = π

Function f (x) = 2cosx ^ 2 + 2 radical sign 3sinxcosx + A, a is a real constant (1) find the minimum positive period of F (x)

f（x）=2cosx^2+2√3sinxcosx+a=cos2x+1+√3sin2x+a=2(sinπ/6cos2x+cosπ/6sin2x)+a+1=2sin(π/6+2x)+a+1
Minimum positive period 2 π / 2 = π

The known function f (x) = 3sinx•cosx+sin2x． (I) find the minimum positive period and monotone increasing interval of function f (x); (II) how can the image of function f (x) be obtained from the image of function y = sin2x?

(1) ∵ function f (x) =
3sinx•cosx+sin2x=
Three
2sin2x+1−cos2x
2=sin(2x−π
6)+1
Two
The minimum positive period of the function f (x) is π (5 points)
By 2K π - π
2≤2x−π
6≤2kπ+π
2(k∈Z)⇒kπ−π
6≤x≤kπ+π
3(k∈Z)，
The monotonic increasing interval of F (x) is [K π − π
6，kπ+π
3]，（k∈Z）．　… (8 points)
（Ⅱ）∵f(x)＝sin(2x−π
6)+1
2＝sin2(x−π
12)+1
2，
ν first shift π from the image of the function y = sin2x to the right
12 units, and then move the image up 1
The graph of function f (x) can be obtained by 2 units (12 points)

The known function f (x) = 3sinx•cosx+sin2x． (I) find the minimum positive period and monotone increasing interval of function f (x); (II) how can the image of function f (x) be obtained from the image of function y = sin2x?

(1) ∵ the function f (x) = 3sinx · cosx + sin2x = 32sin2x + 1 − cos2x2 = sin (2x − π 6) + 12 ᙽ the minimum positive period of the function f (x) is π (5 points) from 2K π − π 2 ≤ 2x − π 6 ≤ 2K π + π 2 (K ∈ z)} K π − π 6 ≤ x ≤ K π + π 3 (K ∈ z), the monotonic increasing interval of ﹣ f (x) is [...]

Find the minimum value of the function FX = 3-2cosx-2sinx + 2 + 2cosx under the root sign

F (x) = under root sign [(1-cosx) ^ 2 + (1-sinx) ^ 2] + under root sign [(0-sinx) ^ 2 + (- 1-cosx) ^ 2]
In this way, finding the minimum value of the function after writing is equivalent to finding the minimum value of the sum of the distances from a point on the unit circle to (1,1) and (- 1,0)
The shortest distance between two points is the distance between two points, which is root five
So the minimum value of F (x) is root five
This paper mainly studies the combination of number and shape

F (x) = - radical 3sin ^ 2x + sinxcosx 1. Ask f (25 schools / 6) 2. Let a belong to (0, PAI), f (A / 2) = 1 / 4-radical 3 / 2, then Sina?

f(x)=-√3（sinx)^2+sinxcosx
=-√3（1-cos2x)/2+sin2x/2
=sin(2x+∏/3)-√3/2
1、f(25∏/6）=sin(25∏/3+∏/3)-√3/2=sin(2∏/3)-√3/2=0
2. F (A / 2) = sin (a + Π / 3) -√ 3 / 2 = 1 / 4 - √ 3 / 2, i.e
sin(a+∏/3)=1/4,0sin(a+∏/3)=1/4<1/2=sin(5∏/6)
A + Π / 3 > 5 Π / 6, that is a > Π / 2
sina+√3cos=1/2,
(sina)^2+(cosa)^2=1
So cosa = (√ 3-3) / 4, Sina = (3 √ 3-1) / 4

F (x) = radical 3sin ^ 2x + sinxcosx to find the minimum positive period

F (x) = radical 3sin ^ 2x + sinxcosx
=(1/2)*sin2x-(√3/2)cos2x+√3/2
=sin(2x-π/3)+√3/2
So the minimum positive period is t = 2 π / 2 = π
If you don't understand, I wish you a happy study!

Find the maximum and minimum of the function y = sin2x-2cos? X

y=sin2x-(1+cos2x)=sin2x-cos2x-1=√2sin(2x-π/4)-1
The maximum value is √ 2-1
The minimum value is - √ 2-1

The minimum value of the function y = 2cos2x + sin2x is______ .

y=2cos2x+sin2x
=1+cos2x+sin2x
=1+
2 (
Two
2cos2x+
Two
2sin2x)
=1+
2sin(2x+π
4)
When 2x + π
4=2kπ−π
2, with a minimum of 1-
Two
So the answer is 1-
Two