The known function f (x) = radical 3sinxcosx + cos ^ 2 x (1) Find the period, range or monotone increasing interval of function f (x) (2) When x [π / 2, π], find the maximum value of function f (x)

The known function f (x) = radical 3sinxcosx + cos ^ 2 x (1) Find the period, range or monotone increasing interval of function f (x) (2) When x [π / 2, π], find the maximum value of function f (x)

F (x) = 1 / 2 * radical 3 * 2sinxcosx + (1-cos2x) / 2 = radical 3 / 2sin2x-1 / 2cos2x + 1 = sin (2x + π / 6) + 1
1、T=2π/2=π +kπ
f(x)∈{0,2}
2X + π / 6 ∈ {- π / 2 + 2K π, π / 2 + 2K π} y
X ∈ {- π / 3 + K π, π / 6 + K π} y
2、
x∈{π/2,π}
2x+π/6∈{π/6+π,π+2π}
sin(2x+π/6)∈{-1/2,1}
The maximum value of F (x) is 2 and the minimum value is 1 / 2

The known function f (x) = cos2x+ 3sinxcosx+1,x∈R. (1) The small positive period and maximum value of F (x) are proved; (2) Find the monotone increasing interval of this function

(1) f (x) = cos2x+
3sinxcosx+1=1
2cos2x+
Three
2sin2x+3
2=sin(2x+π
6)+3
Two
The period of function T = 2 π
2=π
∵-1≤sin(2x+π
6)≤1
∴1
2≤sin(2x+π
6)+3
2≤5
2 is 1
2≤f(x)≤5
Two
(2) When - π
2+2kπ≤2x+π
6≤π
2+2kπ⇒x∈[-π
3+kπ,π
6 + K π] is the monotone increasing interval of function

The known function f (x) = 3 sin square x + 2 root sign 3 SiN x cosx + cos square x, X belongs to R (1) The maximum value and monotone increasing interval of function f (x) are obtained; (2) Let f (x) be greater than or equal to 3

(x) = 3sin2x + 2 √ 3sinxcos x + cos (x = 3 (1-cos2x) / 2 + 3 + sin2x + (1 + cos2x) / 2 = √ 3sin2x + (1 + cos2x) / 2 = 2 = 2 [sin2x * cos (π / 6) - cos2x * sin (π / 6)] + 2 = 2 = 2Sin (2x - π / 6) + 2 (1) when 2x - π / 6 = 2K π + π / 2, K ∈ Z, when 2x - π / 6 = 2K π + π / 2, K ∈ Z, there is a maximum value 4, increase the interval of 4, increase the interval 2, 2, 2, 2, 2, the maximum value 4 increases the interval, the maximum value 4 increases the interval K π - π /

The function f (x) = - 2 root sign 3sin square x + sin2x + root sign 3 is transformed into F (x) = asin (Wx)+

Note (I use sqrt () for square root and PI for PI)
Original formula = sqrt (3) * (1-2 * (SiN x) ^ 2) + sin 2x
=sqrt(3)*cos 2x+sin 2x
=2sin(2x+pi/6)

F {x} = 2sinxcosx + 2 times the root sign 3 of cosx {1} Find the monotone increasing interval {2} of F {x} find the maximum value of F {x} and the set when f {x} takes the maximum value: point out the minimum positive period of F {x}

F {x} = 2sinxcosx + 2 times the root sign 3 of cosx
=sin2x+√3cos2x
=2sin(2x+π/3)
1.2kπ-π/2

Find the minimum positive period of sin 2x + 2 radical sign 3sin? X

sin(2x)+2√3sin²x
=sin(2x)+√3×[1-cos(2x)]
=sin(2x)-√3cos(2x)+√3
=2sin(2x-π/3)+√3
The minimum positive period is 2 π / 2 = π

It is known that f (x) = - 2sinxcosx + 2 times the root sign 3cos square X - the smallest positive period and monotone interval of radical 3, Find the maximum and minimum value of F (x) on [O, Pai / 2], and find the corresponding value of X when the maximum value is obtained, and find the value set of the force that f (x) > is equal to 1

(x) = - 2 SiNx cosx + 2 √ 3cos ^ 2x - √ 3 = - sin2x + 3 (2cos ^ 2x-1) = - sin2x + √ 3cos 2x = - 2Sin (2x - π / 3) minimum positive period: 2 π / 2 = π 2x π / 3 = - π / 2 = π 2x π / 3 = - π / 2 + 2 + 2K π (K ∈ z) x = - π / 12 + K π 2x 2x π / 3 = π / 2 + 2K π (K ∈ z) x = 5 π / 12 + 12 + 12 + 12 + 12 + 12 + 12 + 12 + 12 + 2 + 2K (K ∈ z) x = 5 π (K ∈ z) k π 2x - π / 3 = 3 π / 2 + 2K π (k

The known function f (x) = 2cos ^ 2x + 2 √ 3sinxcosx. (1) find the value range of function f (x) on [- π / 6, π / 3]; and in the supplement (2) In △ ABC, if f (c) = 2,2 SINB = cos (A-C) - cos (a + C), find the value of Tana

f(x)= cos2x+1+√3sin2x
= 2sin(π/6+2x) +1
Range (0,3)
c = π/3
From the following formula 2sinb = 2sinasinc
√3sina = 2sin b
a + b = 120 °
tan a = √3/(√3-1)

Function y = 2cos (2x - π / 6) range

This function is a cos function, so the range is [- 2,2]

It is known that f (x) = 2 times radical 3 + sinxcosx + 2cos? X 1. Find the period of F (x); 2. Monotonically increasing interval of F (x); 3. The maximum value of F (x) and the value range of X to the maximum value

F (x) = 2 times root 3 * sinxcosx + 2cos? X
=√3sin2x+cos2x+1
=2sin(2x+π/6) +1
(1) T=2π/2=π
(2) 2kπ-π/2≤2x+π/6≤2kπ+π/2
2kπ-2π/3≤2x≤2kπ+π/3
kπ-π/3≤x≤kπ+π/6
Increasing interval [K π - π / 3, K π + π / 6], K ∈ Z
(3) When π + π + π + π / π = 2 x + π, the maximum value is obtained