f x 2 root number 3asinxcosx 2asinx sinx b a 0 1 Find the period and maximum value of f x and write

1)f(x)=2√3 asinxcosx-2a(sinx)² b (a>0, b∈r)

3 asin2x-2a* (1-cos2x)/2 +b

3 asin2x + acos2x +b-a

2a(√3/2 sin2x + 1/2 cos2x) +b-a

2a sin(2x+∏/6) +b-a

So t=2 2= ,a>0, f(x)max=2a+b-a=a+b

When 2x+6=2k+2, f(x) is the maximum, and x=k + 6 (k z).

2) x [0, 2], then 6 2x+ 6 7 6, so -1 2 sin(2x+ 6) 1

So b-2a f(x)=2a sin(2x+ 6) +b-a b+a, known -1 f(x) 8

i.e. b-2a=-1, b+a=8, a=3, b=5

Note: The answer on the second floor is basically wrong! Mistaken !!

Because f(x)=2 root number 3asinxcosx-2asinx*sinx+b(a 0).

Root number 3*a*sin2x - a*(1-cos2x) +b root number 3*a*sin2x + a*cos2x -a+ba*[2*sin(2x+ 6)]-a+b, so period t=2 2=

Because sin(2x+ 6) [1,1].

So the maximum value of f(x) is 2a-a+b=a+b, and x=6+k.

2] f(x) defines the domain as [0, 2] and the value range is [-1,8] to find a, and b because the defined domain is [0, 2].

So sin(2x+6).

Substituting f(x) [b,a+b].

So b-2a=-1, a+b=8

So a=3, b=5

There is only one question: f(x)=root number 3asin2x-a acos2x b, f(x)=2asin(2x pie 6) b-a, so period t=pie, f(x) is maximum a b

f(x)=2 root number 3sin square mill closed x + sin2x + root number 3 root number 3 (blind crack 2sin square x 1) 晌caution sin2x root number 3 (1 cos2x 1) sin2x 2 root number 3 root number 3 cos2x sin2x 2 sin (2x 60 degrees) 2 root number 3 so t 360 degrees w 180 degrees and because of sin(2x ..

f（x）=2√3cos²x-sin2x-√3=√3（2cos²x-1）-sin2x

3cos2x-sin2x

2（cos2xsinπ/3-sin2xcosπ/3）=2sin（π/3-2x）

So the minimum positive period of the function k=2 2=

Minimum = -2

f(x)=2√3cos^2x-2sinxcosx-√3=√3(cos2x+1)-sin2x-√3=√3cos2x-sin2x

2(sin2xcosπ/3-cos2xsinπ/3)=-2sin(2x-π/3)

Minimum positive period =

The minimum value -2 monotonically increases the interval: x (k +5 12, k +11 12) where k z

f(x) = 2 root number 3cos 2x - 2sinxcosx - root number 3 = 3 (1 + cos2x) - sin2x - 3 = 3 cos2x - sin2x

2sin(2x-π/3)

The minimum positive period of the function is

The minimum value is -2

The monotonically increasing interval is [k -7 12, k - 12] (k z).

It is known that the source pin fibrillation function f(x) = 2 3sinxcosx + 2cos x - 1

Find the value of f(6) and the minimum positive period of f(x).

f(x) =2√3sinxcosx + 2cos²x - 1√3sin2x + 1 + cos2x) -1√3sin2x + cos2x

3 + 1) sin(2x + 30°) acosx + bsinx = a² +b²) sin(2x +

2sin(2x + 30°)

f(π/6) =2 sin60° =3

The minimum positive period is 2 and the hail is defeated with 2 =

The heart is no longer cold, hello.

f(x)=3sinx 2+2 root number 3sinxcosx + 5cosx 2=3* (1-cos2x) 2 + root number 3sin2x+5* (1+cos2x) 2

3 2-3 2cos2x + root number 3sin2x+5 2+5 2cos2x

4 + cos2x + root number 3sin2x

4+2sin（2x+π/6）

t=2π/2=π

The maximum value = 6f(a) = 5, i.e. sin(2x+6)=1 2 a=0

So tana=tan0=0

f(x)=sinxcosx-√3cos^2x+√3/2.So f(x)= 1 2sin2x- 3 2(1+cos2x)+ 3 2=1 2sin2x- 3 2cos2x=sin(2x- 3), so doing a small positive period is 2 2=. The maximum value is 1, and when the maximum value is obtained, 2x- 3= 2+2k

So x=5 12+k, note that k is an integer, and is written as a set.

f(x)=√2sin2x+√2cos2x

2[√2/2sin2x+√2/2cos2x]=2sin(2x+π/4)

The minimum positive period is: t=2 2=

When 2x+ 4= 2+2k

That is, when x= 8+k, f(x) is the maximum value and f(max)=2

Solution: f(x) = 2·sin2x + 2·cos2x

Let tan = 2 2=1, then take = 4, f(x)= [(2) +2) ]sin(2x+ 4)=2sin(2x+ 4).

The maximum value of f(x) = 2, and the minimum positive period = 2 |2|=π

By the equation of the doubling angle: f(x)=sin2x- 3cos2x by the auxiliary angle formula: f(x)=2sin(2x- 3) Therefore, the minimum positive source eggplant period t=2 2=

The maximum value is 2, so you have fun! I hope it can help you, if you don't understand it, please ask, I wish you progress in learning! o(∩_o