Knowing the function f X negative root number 3sin square x sinxcosx, find the minimum positive peri

f(x) = negative root number 3sin square x + sinxcosx root number 3 2 - root number 3 2 cos2x + 1 2sin2xsin (2x-3) + root number 3 2

So, the cycle is

f(x) = -3^(1/2)[sin(x)]^2 + sin(x)cos(x)

3^(1/2)[1-cos(2x)]/2 + sin(2x)/2

sin(π/3)cos(2x) +cos(π/3)sin(2x) -3^(1/2)/2

sin(2x + /3) -3^(1/2)/2

2x + 2 + 3 = 2(x + 3, so the minimum positive period of the function f(x) = -3 (1 2)[sin(x)] 2 + sin(x)cos(x) is .

Also, y(x) = x 2 + 2x + 1,y'(x) = 2x + 2 = 2(x+1)

y'(1) = 2(1+1) = 4.

So, y(x) = x 2 + 2x + 1 is equal to 4

Root number 3 (1-cos2x) 2 + 1 2sin2x = sin(2x + 3) - root number 3 2

t=2π/2=π

y'=2x+2, so the result is 4

f(x)=root3*sin 2x6)-1 hail2 =2 so t=2 or hail =

f(x)=(3 2)·2sinxcosx (1+cos2x) 2

3 2)sin2x (1 2)cos2x 1 size imitation 2cos( 6)sin2x sin( 6)cos2x 1 pants 2sin(2x 6) 1 2

The minimum positive period of f(x) t=2 2=

f(x)=√3sinx+cosx

2[sinxcosπ/6+cosxsinπ/6]2sin(x+π/6)

The minimum banquet talks about the Zhou Tangerine period t=2 |w|=2 1=2 Note: w is the number of round auspicious beams of x!

This question may be more complicated if it is denaturalized.

So use the analytic method.

1) In the defined domain, the period of sin squared x is obviously the root number 3sinxcosx = root number , so the period is also ;

In summary, the minimum period of the function f(x) is

2)(sinx) 2 >=0, the root number 3sinxcosx >=0 So guess that the minimum value of f(x) is 0, and we now find out if there is such an x due to (sinx) 2 >=0, the root number 3sinxcosx >=0, and f(x)=sin square x + root number 3sinxcosx=0

So only (sinx) 2 = 0, the root number 3sinxcosx =0 is not difficult to find, if (sinx) 2 = 0, then x=k If the root number 3sinxcosx = 0, then x=sum up, so that the function takes the minimum value of x **x=k k k belongs to the integer, the minimum value of f(x) = 0,6, one ,2,f(x) =1 2 - cos2x) 2 + 3sin2x 2

1/2 + sin(2x - 6）

The minimum period of f(x) =

The minimum value of x is **

k +5 6,1, original simplification f(x) = sinx (sinx + root number 3sinxcosx) with the formula of the sum of two angles and the difference is reduced to f(x) = 2sinxcos (丌 6 -x) The sum difference of the product is sin (2x-6 parts of the same) + the root number of 2 points is 3 The minimum positive period is the minimum value of the root number of 2 points 3 minus 1 ** is 3 points of the root number of 3 points minus k to 3 points of the number + k of the difference This is a regular problem in high school mathematics in terms of trigonometric functions and properties Find a few example problems to see A lot of them are in the set of formula concepts, mobile phone party, tired to death, give some points. 0, known function f(x) = sin square x + root number 3sinxcosx, (1) find the minimum period of the function f(x).

2) Find the minimum value of the function f(x) and write the ** that makes the function wild swim to take the minimum value of x

f(x)=sinx-root number 3cosx

2 (1 2sinx-root number or number 3 2cosx) 2 (sinxcos 3-cosx sin 3) 2sin(x-3).

Minimum Positive Cycle Wisdom Period: 2

f(x) = root number 2 (sinx + cosx) = root number 2 * root number 2 (sinxcos45 + sin45cosx) = 2sin (x+45).

Therefore, the minimum positive period of the shouting faction is t=2 and the infiltration is 1=2

f(x)=sinx-root number 3cosx

2 (reed with 1 2sinx-root number cheating first rent number 3 2cosx) 2 (sinxcos 3-cosx sin 3) 2sin(x- 3).

Minimum positive period: 2

Solution: f(x) = 3

sinxcosx-cos x= 3 2sin2x-(cos2x+1 2)= 3 2sin2x-1 2·cos2x-1 2 sin 2=sin2xcos 6-sin 6cos2x-1 2=sin(2x- 6)-1 2

Therefore, the minimum positive circumference is t=2 2= hope.