Given the function f (x) = SiNx + radical 3cosx, R simplifies f (x) and finds its period; finds the maximum value of F (x) and the corresponding value of X at this time; finds f (x) The known function f (x) = SiNx + radical 3cosx, R Simplify f (x) and find its period; Find the maximum value of F (x) and the corresponding value of X at this time; Find the monotone increasing interval of F (x);

f(x)=2*(1/2)sinx+√3/2cosx
=2sin(x+π/6)
T=2π/1=2π
f(x)max=2
In this case, x + π / 6 = k π + (π / 2)
X = k π + (π / 3) k ∈ integer
Increase range:
kπ-（π/2）≤x+π/6≤kπ+（π/2）
kπ-(2π/3) ≤x≤kπ+(π/3)

When - 2 parts π ≤ x ≤ 2 parts π, the value range of function y = SiNx + root 3cosx

Y = SiNx + radical 3cosx
=2[sinx*(1/2)+cosx*(√3/2)]
=2[sinx*cos(π/3)+cosx*sin(π/3)]
=2sin(x+π/3)
∵ - 2 parts π ≤ x ≤ 2 parts π
∴ -π/6≤x+π/3≤5π/6
∴ sin(x+π/3)∈[-1/2,1]
∴ 2sin(x+π/3)∈[-1,2]
That is, the value range of the function y = SiNx + radical 3cosx is [- 1,2]

Given the vector a = (root 3cosx, 0), B = (0, SiNx), write the function f (x) = (a + b) ^ 2 + Radix 3sin2x) (1) find the minimum value of function f (x) and take the minimum value (2) if the image of function f (x) is shifted by vector, the image obtained is centrosymmetric with respect to the origin of the coordinate, and decreases monotonically on [0, π / 4], then the minimum length d is obtained

Known vector a = (root 3cosx, 0), B = (0, SiNx), note the function f (x) = (a, b) = (a, b) = (a, b) = (a, b) = (a, b) = (a, b) = (a, b) = (a, b) = (a, b) = (a, b) = (a, b) = ( I went to drink milk tea (√ 2: process

What is the minimum value of the function f [x] = SiNx + radical 3cosx?

F(x)=2sin(x +π/3)
The minimum value is - 2

Function f (x) = radical 3cosx SiNx (0

F (x) = 2 (sinwu / 3cosx coswu / 3sinx)
=2Sin (Wu / 3-x)
Zero

When - Pai / 2 is less than or equal to X and less than or equal to Pai / 2, the value range of the function f (x) = SiNx + radical 3cosx is Completion

Function y = 2 (1 / 2sinx + radical 3 / 2 cosx)
=2sin（x+π/3）
Because - π / 2 ≤ x ≤ π / 2,
-π/6≤x+π/3≤5π/6,
Therefore, - 1 / 2 ≤ sin (x + π / 3) ≤ 1, function range [- 1,2]

Find the function f (x) = cos square x + root sign 3sinx times cosx Find the maximum and minimum values

Cox ^ 2 + root sign 3sinx times cosx
=0.5 (1 + cos2x) + radical 1.5 * sin2x
Derivative is enough

Y = maximum and minimum of sin2x / (SiNx + cosx + 1)

(sinx+cosx)²=sin²x+cos²x+2sinxcosx
=1 + sin2x < = 1 + 1 = 2, if and only if sin2x = 1, i.e. x = 45 °, the equal sign holds
So SiNx + Cox < = radical 2
That is, y = SiNx + cosx + 1 < = 1 + radical 2
So the maximum value of the function y = SiNx + cosx + 1 is 1 + radical 2

The maximum value of y = 3 | SiNx | + 4 | cosx |

If SiNx and cosx are all greater than 0, then it will be easy to handle
y=5sin(a+ x)
→ymax=5
ok
Y = 3sinx + 4cosx = 5 * (3 / 5 * SiNx + 4 / 5 * cosx) = 5 * sin (x + a), and a refers to the relevant angles of a right triangle surrounded by 3,4,5 sides

Find the maximum value of y = sin2x - 3 (SiNx + cosx) ..

Let a = SiNx + cosx = √ 2Sin (x + π / 4)
So - √ 2