F (x) = - sin2x + 2 times root sign 3sin? X - radical sign 3 + 1, find the minimum positive period and the simple decreasing interval

F (x) = - sin2x + 2 times root sign 3sin? X - radical sign 3 + 1, find the minimum positive period and the simple decreasing interval

(x) = - sin (2x) + 2 √ 3sin (2x) + 2 √ 3 (3 + 3 + 1 = - sin (2x) + 3 [1-cos (2x)] - - 3 + 3 + 1 = - sin (2x) - √ 3cos (2x) + 1 = (- 2) [(1 / 2) sin (2x) + (√ 3 / 2) cos (2x)] + 1 = (- 2) sin (2x + π / 3) + 1 minimum positive cycle (Tmin = 2 π / 2 = π / 2 = π (2x + π / 3) and 3) monotone monotone monotone monotone monotone and monotone monotone monotone monotone monotone and monotone monotone monotone monotone monotone monotone and monotone monotone monotone monotone monotone monotone when increasing, - 2Sin (2x

Given the function f (x) = √ 3sin ω xcos ω x + cos 2 ω x, X ∈ R, ω > 0, (1) find the value range of function f (x); (2) if the minimum positive period of function f (x) is π / 2, then when x ∈ [0, π / 2], we can find the monotone decreasing interval of F (x) For detailed explanation, steps are required. Thank you

f(x)=√3sinωxcosωx+cos²ωx
=√3/2*2sinωxcosωx+cos²ωx
=√ 3 / 2 * sin2 ω x + (Cos2 ω x + 1) / 2
=√3/2*sin2ωx+1/2*cos2ωx+1/2
=sin(2ωx+π/6)+1/2
Max = 1 + 1 / 2 = 3 / 2
Minimum = - 1 + 1 / 2 = - 1 / 2
The range is [- 1 / 2,3 / 2]
(2) The minimum positive period is π / 2 = 2 π / 2 ω
∴ω=2
f(x)=sin(4x+π/6)+1/2
x∈[0,π/2]
4x+π/6∈[π/6,13π/6]
SiNx is a decreasing function on [π / 2,3 π / 2]
∴4x+π/6∈[π/2,3π/2]
x∈[π/12,π/3]
The f (x) minus interval is [π / 12, π / 3]

Given that the function f (x) = 2 √ 3sinxcosx-3sin? X-cos? X + 2 (1), when x ∈ [0, π / 2], find the range of F (x)

f(x)=2√3sinxcosx-3sin²x-cos²x+2=√3sin2x-3(1-cos²x)-cos²x+2=√3sin2x-3+3cos²x-cos²x+2=√3sin2x+2cos²x-1=√3sin2x+cos2x=2(√3/2*sin2x+1/2*cos2x)=2sin(2x+π/6）∵x∈[...

It is known that the minimum positive period of the function y = √ 3sin ω xcos ω x-cos 2 ω x + 3 / 2 (x ∈ R, ω∈ R) is π, and when π / 6, the function has a minimum value (1) Find the analytic formula of function f (x) (2) Finding monotone increasing interval of function f (x)

（1）f（x）=√3sinωxcosωx-cos²ωx+3/2
=√3/2sin2ωx-(1+cos2ωx)/2+3/2
=√3/2sin2ωx-1/2cos2ωx/2+1
=sin(2ωx+π/4)+1
∵ the minimum positive period is π
∴T=2π/ω=π
∴ω=1
∴f（x）=sin(2x+π/4)+1
(2) Forget to let me know if anyone does it

Given the function f (x) = sin ^ 3xcosx + cos ^ 3xsinx + √ 3sin ^ 2x, find the monotone decreasing interval of the function and find y = (x) (0 ≤ x ≤ π) ）Value range of

f(x)=sin³xcosx+cos³xsinx+√3 sin²x
=sinxcosx(sin²x+cos²x)+√3 (1-cos2x)/2
=½ sin2x - √3 /2 cos2x + √3 /2
=sin(2x-π/3) + √3 /2
The decreasing interval of F (x) is 2x - π / 3 ∈ [π / 2 + K π, 3 π / 2 + K π] (K ∈ z)
That is, X ∈ [5 π / 12 + K π / 2,11 π / 12 + K π / 2] (K ∈ z)
The monotone decreasing interval in 0 ≤ x ≤ π is x ∈ [5 π / 12,11 π / 12]
The range is [√ 3 / 2 - 1, √ 3 / 2 + 1]

The image crossing point (0, - radical 3) of the function f (x) = 2asin? X + 2sinxcosx-a is known (1) If x belongs to [0, π / 2], find the range of function f (x)

f(0)=0+0-a=-√3
a=√3
f(x)=2√3(1-cos2x)/2+sin2x-√3
=sin2x-√3cos2x
=2sin(2x-π/3)
-π/3<=2x-π/3<=2π/3
So the maximum is 2 sin π / 2 = 2
The minimum is 2Sin (- π / 3) = - √ 3
So the range is [- √ 3,2]

Given vector a = (SiNx, - 1) vector b = (root 3cosx, - 1 / 2), function f (x) = (vector a + vector b) * vector A-2 Let a, B, C be the opposite sides of the inner angles a, B and C of the triangle ABC, where a is an acute angle, a = 2, radical 3, C = 4, and f (a) = 1, find the area s of a, B and triangle ABC

Vector a = (SiNx, - 1), vector b = ((√ 3) cosx, - 1 / 2), function f (x) = (a + b) · A-2;
It is known that a, B and C are the opposite sides of a, B and C, respectively, where a is an acute angle, a = 2 √ 3, C = 4, and f (a) = 1,
Find the area s of a, B and triangle ABC
a+b=(sinx+(√3)cosx,-1-1/2)=(sinx+(√3)cosx,-3/2)；
So f (x) = (a + b) · A-2 = [SiNx + (√ 3) cosx] SiNx + 3 / 2-2 = sin? 2x + (√ 3) sinxcosx-1 / 2
=(1-cos2x)/2+(√3/2)sin2x-1/2=(√3/2)sin2x-(1/2)cos2x=sin2xcos(π/6)-cos2xsin(π/6)
=sin(2x-π/6)
Since f (a) = sin (2A - π / 6) = 1, 2A - π / 6 = π / 2, 2A = π / 2 + π / 6 = 2 π / 3,  a = π / 3
From the cosine theorem, there is a  2 = B  2 + C  2 - 2bccosa, substituting the known value 12 = B  2 + 16-4b, that is, B  4B + 4 = (b-2) 2 = 0, so B = 2;
SΔABC=(1/2)bcsinA=(1/2)×2×4×sin(π/3)=2√3.

Given the vector a = (SiN x, - 1) B = (root 3cos x, - 1 / 2), the function f (x) = (a + b) A-2 Finding the minimum period of positive function T (f)

f(x)=a²+ab-2=(sin²x+1)+(√3sinxcosx+1/2)-2
=√3/2sin2x-1/2cos2x
=sin(2x-π/6)
Minimum positive period T = 2 π / 2 = π
The minimum positive period of (x) is π

Given the vector a = (2cos (x / 2), Tan (x / 2 + π / 4)), B = (Radix 2Sin (x / 2 + π / 4), Tan (x / 2 - π / 4)) let FX Given vector a = (2cos (x / 2), Tan (x / 2 + π / 4)), B = (Radix 2Sin (x / 2 + π / 4), Tan (x / 2 - π / 4)), Let f (x) = a × B, Find the maximum and minimum positive period of function f (x), and write the monotone interval of F (x) on [0, π]

(x) = 2cosx / 2 × ((x / 2 + π / 4) + Tan (x / 2 + π / 4) + Tan (x / 2 + π / 4) × Tan (x / 2 - π / 4)) = √ 2 [sin (x + π / 4) + sin (π / 4)) + [1 + Tan (x / 2)] / [1-tan (x / 2)] / [1-tan (x / 2)] / [Tan (x / 2)] / [1 + Tan (x / 2)] = √ 2Sin (x + π / 4) maximum value = √ 2, minimum positive cycle = 2 π, 2 π = 2 π = 2 π 2 π = 2 π = 2 π = 2 π = 2 π = 2 π = 2 π = 2 π = 2 π the increasing region of SiNx

Given the vector a = (2cosx / 2,1 + Tan ^ 2x), B = (Radix 2Sin (π / 4 + X / 2), cos ^ 2x), Let f (x) = a * B 1 find the monotone interval of F (x) on [0, π / 2] 2 if f (a) = 11 / 4 and a belongs to (π / 2, π), find the value of F (- a)

According to the vector operation rule, f (x) = 2 cosx / 2 * Radix 2Sin (π / 4 + X / 2) + cos ^ 2x + Tan ^ 2x * cos ^ 2x = 2 * Radix 2 * sin (π / 4 + X / 2) * cosx / 2 + cos ^ 2x + sin ^ 2 x = 2 * Radix 2 * 1 / 2 * [sin (x + π / 4) + sin (π / 4)] + 1 = radical 2 * sin (x + π / 4) +