Given that the function f (x) = 2A * cos ^ 2 x + b * SiN x * cos x satisfies f (0) = 2, f (π / 3) = 1 / 2 + (radical 3) / 2, find the maximum and minimum value of F (x); If α and β are greater than 0 and less than π, f (α) = f (β), and α is not equal to β, find the value of Tan (α + β)

Given that the function f (x) = 2A * cos ^ 2 x + b * SiN x * cos x satisfies f (0) = 2, f (π / 3) = 1 / 2 + (radical 3) / 2, find the maximum and minimum value of F (x); If α and β are greater than 0 and less than π, f (α) = f (β), and α is not equal to β, find the value of Tan (α + β)

(1) The meaning of the title
f(0)=2a=2 f(π/3)=1/2a+√3/4b=1/2+√3/2
∴a=1,b=2
∴f(x)=2cos²x+2sinxcosx=1+cos2x+sin2x=1+√2sin(π/4+2x)
The maximum value of F (x) is 1 + √ 2, and the minimum value is 1 - √ 2
（2）f(a)=f(b)∴1+√2sin(π/4+2a)=1+√2sin(π/4+2b)
sin(π/4+2a)=sin(π/4+2b)
∵a≠b
ν π / 4 + 2A + π / 4 + 2B = 2 π or π
A + B = π / 4 or 3 π / 4
∴tan(a+b)=±1

The minimum positive period of the function f (x) = root 3sinx + 3cosx is A. Pie / 2 b.pie C.2 pie d.pie We need to go through it in detail Why is C? Can you elaborate I will be very grateful

F (x) = Radix 3sinx + 3cosx = 2 √ 3 (SiNx / 2 + √ 3 / 2 * cosx) = 2 √ 3sin (x + π / 3) t = 2 π / w = 2 π select C

The function y = 3sinx radical 3cosx belongs to the range of [0, Pai] at X

y=3sinx-√3cosx
=2√3(sinx*√3/2-cosx*1/2)
=2√3(sinx*cosπ/6-cosx*sinπ/6)
=2√3sin(x-π/6)
∵x∈［0,π］
∴x-π/6∈［-π/6,5π/6］
When X - π / 6 = - π / 6, the minimum value of 2 √ 3 × (- 1 / 2) = - √ 3 is obtained
When X - π / 6 = π / 2, the maximum value of 2 √ 3 × 1 = 2 √ 3 is obtained
So the range is [- √ 3,2 √ 3]

If y = 3sinx-3, why is the maximum value of 3cosx 6?

y=6sin(x-π/3),x∈R
So the maximum is 6

What is the minimum value of the function y = 2 root sign 3sinx + 2cosx radical 2

Y=2√3sinX+2cosX-√2
=4(sinx*√3/2+cosx*1/2)-√2
=4sin(x+ π/6)- √2
So the minimum value of Y is - 4 - √ 2 and the maximum value is 4 - √ 2

Let f (x) = radical 3 / 2sinx + 1 / 2cosx. Find the minimum positive period and the maximum and minimum of the function

The original formula = cos no / 6sinx + Sinn / 6cosx
=Sin (n / 6 + x)
T = 2 N / 1 = 2 n
f(X)max=1
f(X)min=-1

What is the minimum value of the function y = 2 root sign 3sinx + 2cosx radical 2

y=2√3 sinx-2cosx-√2
=4(√3/2 sinx-1/2cosx)-√2
=4(sinx cosπ/6- cosx sinπ/6)-√2
=4sin(x-π/6)-√2
When sin (x - π / 6) = - 1, y = - 4 - √ 2

What is the minimum value of the function f (x) = cosx + Radix 3sinx (0 ＜ = x ＜ = π / 2) f (x)?

f(x)=√3sinx+cosx
=2(√3/2sinx+1/2cosx)
=2sin(x+π/6)
0≤x≤π/2
π/6≤x≤2π/3
Sin t increases first and then decreases at π / 6 ≤ x ≤ 2 π / 3 [],
1/2≤sin(x+π/6)≤1
1≤2sin(x+π/6)≤2
f(min)=1

The function f (x) = sin (x + 4) - root sign 3cos (x + 4), find the maximum value of function f (x)

f(x)=2(1/2sin(x+4)-v3/2cos(x+4))
=2sin(x+4-π/6)
The maximum is 2

The function f (x) = 3sinx+sin(π The maximum value of 2 + x) is______ .

By F (x) =
3sinx+cosx＝2sin(x+π
6)⇒f(x)max＝2．
So the answer is: 2