The minimum value and minimum positive period of the function FX = sinxcosx are

The minimum value and minimum positive period of the function FX = sinxcosx are

f(x)=(2sinxcosx)/2
=(sin2x)/2
The period is 2 π / 2 = π
When the minimum value is - 1 / 2, sin2x = - 1

If the function FX = sinxcosx, then the maximum value and minimum positive period of FX are

f(x)=sinxcosx=1/2sin2x
So the maximum value is 1 / 2 * 1 = 1 / 2
The minimum value is 1 / 2 * (- 1) = - 1 / 2
T=2π/2=π

Let f (x) = - 1 + 2 radical sign 3sinxcosx + 2cos2x. Find the function y = f (x), X belongs to the maximum value of {- 7 π / 12,5 π / 12} and the corresponding value of X Urgent

f(x)=-1+√3sin2x+2cos2x
=-1+√7sin(2x+t)
Think about it. Is the coefficient 2 in front of cos2x redundant?

Given the function f (x) = root sign 3sinxcosx-1 / 2cos2x 1, if x belongs to [0, Pai / 2], it is to find the maximum value of the function

F (x) = Radix 3sinxcosx-1 / 2cos2x
=√3/2sin2x-1/2cos2x
=sin（2x-π/6）
When 2x - π / 6 = π / 2, that is, when x = π / 3, the function gets the maximum value
The maximum value is 1

What is the maximum value of the function SiNx square + root sign 3sinxcosx in the interval [π in 4, π in 2]

Y=sin²x+√3sinxcosx
=(1-cos(2x))/2+√3/2*sin(2x)
=√3/2*sin(2x) -cos(2x)/2+1/2
= sin(2x-π/6) +1/2
X∈[π/4, π/2],
2x-π/6∈[π/3, 5π/6],
When 2x - π / 6 = 5 π / 6, the minimum value of the function is 1 / 2 + 1 / 2 = 1
When 2x - π / 6 = π / 2, the maximum value of the function is 1 + 1 / 2 = 3 / 2

The function f (x) = 3sin square x + 2 root sign 3sinxcosx + 5cos square x is known Find the period and maximum of function f (x) Given f (a) = 5, find the value of Tana

I just answered the question:
(I)
f(x)=3sin^2x+2√3sinxcosx+5cos^2x
=sin^2 x+2√3sinxcosx+3cos^2 x+2(sin^2 x+cos^2 x)
=sin^2 x+2√3sinxcosx+(√3)^2·cos^2 x+2
=(sinx+√3cosx)^2+2
=[√(1^2+(√3)^2)·sin(x+arctan(√3/1)]^2+2
=[2sin(x+π/3)]^2+2
=4·sin^2 (x+π/3) +2
∵sin(x+π/3)≤1
∴sin^2 (x+π/3)≤1
ν f (x) = 4 · sin ^ 2 (x + π / 3) + 2 ≤ 6; the maximum value is 6
f(x)=4·sin^2 (x+π/3) +2
=2·[2sin^2 (x+π/3)-1] +2 +2
=-2·[1-2sin^2 (x+π/3)]+4
=-2cos(2x+3π/3)+4
Then the minimum positive period of F (x) is 2 π / 2 = π
(II)
Substituting the original formula:
f(a)=3sin^2 a+2√3sinacosa+5cos^2 a
=5;
∴3sin^2 a+2√3sinacosa=5(1-cos^2 a)=5sin^2 a;
2√3sinacosa=2sin^2 a;
√3sinacosa=sin^2 a;
Divide both sides by sin ^ 2 a at the same time
√3/tana=1;
∴tana=√3.

F (x) = Radix 3sin2x-2sin square x (1). Find the maximum value of F (x) (2). Find the set of zeros of F (x)

F (x) = (√ 3) sin2x-2sin? X; (1) find the maximum value of F (x) (2) find the set of zeros of F (x)
(1)f(x)=(√3)sin2x+cos2x-1=2[(√3/2)sin2x+(1/2)cos2x]-1=2[sin2xcos(π/6)+cos2xsin(π/6)]-1
=2sin(2x+π/6)-1
-2≦2sin(2X+π/6)≦2,-3≦2sin(2x+π/6)-1≦1
That is - 3 ≤ f (x) ≤ 1, so the maximum value of F (x) is 1 and the minimum value is - 3
(2) Let f (x) = 2Sin (2x + π / 6) - 1 = 0
Sin (2x + π / 6) = 1 / 2, so 2x + π / 6 = π / 6 + 2K π, x = k π, K ∈ Z

Known function f (x)= 3sin2x-2sin2x． (I) find the maximum value of function f (x); (II) find the set of zeros of function f (x)

(I) ∵ f (x) = 3sin2x-2sin2x = 3sin2x + cos2x-1 = 2Sin (2x + π 6) - 1, so the maximum value of the function f (x) is 2-1 = 1 (II) from F (x) = 0, 23sin xcos x = 2sin2x, then SiN x = 0, or 3cos x = SiN x, namely Tan x = 3, x = k π from sin x = 0, x = k π from Tan x = 3

Let f (x) = 2cos square x + root sign sin2x, find the set of zeros of function f (x)

f=2cos^2x+√sin2x
Because cos ^ 2x ≥ 0, √ sin2x ≥ 0, it can only be equal to 0 if both are 0
Cos ^ 2x = 0 means x = k π + π / 2
Sin2x = 0 means x = k π / 2
Therefore, the common part is x = k π + π / 2, which is the zero point

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) When f (x) = 3sin2x + cos2x = 2 (sin2xcos π 6 + cos2xsin π 6) = 2Sin (2x + π 6)  t = 2 π 2 = π, when 2x + π 6 = 2K π + π 2, K ∈ Z, the function obtains the maximum value 2