Given the function FX = 2 sinxcosx + cos2x, find the minimum positive period and maximum value of FX?

Given the function FX = 2 sinxcosx + cos2x, find the minimum positive period and maximum value of FX?

f(x)=sin2x+cos2x=√2sin(2x+π/4)
Minimum positive period T = 2 π / 2 = π
The maximum value is √ 2

FX = - √ 3sin squared x + sinxcosx, find the value of F (Wu / 6) and find the minimum positive period and maximum value of function FX FX = - √ 3sin squared x + sinxcosx, find the value of F (Wu / 6) and find the minimum positive period and maximum value of function FX

If (x) = - (3 / 2) (1-cos2x) + (1 / 2) sin2x = (1 / 2) sin2x = √ 3 / 2 cos2x + 1 / 2 sin2x - √ 3 / 2 cos2x + 1 / 2 sin2x - √ 3 / 2 = sin (2x + π / 3) - √ 3 / 2 / 2F (Wu / 6) = sin2 π / 3 / 3 / 3 / 2 = 3 / 2 / 3 / 2 = 0 cycle T = 2 π / 2 = π when sin (2x + π / 3) = 1, the maximum value of F (x) is the maximum value of F (x) when sin (2x + π / 3) = 1, the maximum value of F (x) is the maximum value of F (x) when sin (2x + π / 3) = 1 - √

Let f (x) = 3 SiN x + 2 times root sign 3 SiN x cos x + 5 cos X X. ask the period and maximum value of function f (x)) Urgent

3sin square x = 3 (1-cos2x) / 2
3 SiNx cosx = 3 times SiNx cosx
5cos square x = 5 (1 + cos2x) / 2
After the three transformations, the sum of F (x) = sin2x + cos2x + 4 is obtained
=2sin(2x+π/6)+4
The period is π
The maximum value is 6 when x is k π + π / 6

The function y = = 3sin square x + 2 root sign 3sinxcosx + 5cos square X

y=3+√3sin2x+2(1+cos2x)/2
=√3sin2x+cos2x+4
=2sin(2x+π/6)+4
So the maximum 2 + 4 = 6
Minimum - 2 + 4 = 2
Range [2,6]

. known radical 2

Four
Just draw a picture
First draw the image of the radical a ^ 2-x ^ 2. This is a semicircle, y is greater than zero, and the radius is a
Draw 2 - | x | again
Because when a circle and a straight line are tangent, the radius is equal to root 2
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The known function f (x) = 2sin2x + 2 times root sign 3sinxcosx + 1 Find the minimum positive period, monotonically increasing interval

F (x) = 2sin2x + 2 times root sign 3sinxcosx + 1
=(2 + radical 3) sin2x + 1
Minimum positive period π, monotonically increasing interval [K π - π / 4, K π + π / 4]

How to find the maximum value of the function f (x) = root 3sin2x-cos2x

"Mathematical Q & a team" for you to answer, I hope to help you
f(x)=√3sin2x-cos2x
=2（√3/2sin2x-1/2cos2x ）
=2sin（2x-π/6 ）
So the maximum value is 2, the minimum value is - 2

Known function f (x)= 3sin2x+cos2x． (1) Find the minimum positive period and maximum value of function f (x); (2) Find the monotone increasing interval of function f (x)

（1）f（x）=
3sin2x+cos2x=2（sin2xcosπ
6+cos2xsinπ
6）=2sin（2x+π
6）
∴T=2π
2=π，
When 2x + π
6=2kπ+π
2, K ∈ Z, i.e. x = π
When 6 + K π, K ∈ Z, the maximum value of the function is 2
When 2x + π
6=2kπ-π
2, that is, x = k π - π
When k ∈ Z, the minimum value of the function is - 2
(2) When 2K π - π
2≤2x+π
6≤2kπ+π
6, K ∈ Z, that is, K π - π
3≤x≤kπ+π
6, K ∈ Z, the function increases monotonically,
The monotonic increasing interval of F (x) is: [K π - π
3，kπ+π
6]，k∈Z．

Function y= The minimum positive period of 3sin2x + cos2x is______ .

Y=
3sin2x+cos2x=2（
Three
2sin2x+1
2cos2x）=2sin（2x+π
6），
∵ω=2，∴T=2π
2=π．
So the answer is: π

Find the amplitude and period of the function y = cos2x + radical 3sin2x

y=cos2x+√3sin2x
=2（1/2cos2x+√3/2sin2x）
=2（sin30°cos2x+cos30°sin2x）
=2（sin2x+30°）
The amplitude is 2
Period T = 2 π / ω = 2 π / 2 = π