Let f (x) = A-B, where vector a = (m, cos2x), B = (1 + sin2x, 1), X belongs to R, and the image of function y = f (x) passes through point {4} π, 2} Find the value of the real number M 2 find the minimum value of function f (x) and the set of values of X at this time

Let f (x) = A-B, where vector a = (m, cos2x), B = (1 + sin2x, 1), X belongs to R, and the image of function y = f (x) passes through point {4} π, 2} Find the value of the real number M 2 find the minimum value of function f (x) and the set of values of X at this time

1: From the meaning of the title,
f(x)=a-b=(m-1-sin2x,cos2x-1)
And f (x) passes through the point (π / 4,2);
∴ m-1-sin2x=π/4 (1)
cos2x-1=2 (2)
The solution is: M = 1 + π / 4;
2: Find the minimum value of function f (x) and the set of values of X at this time

The vector a = (m, sin2x), B = (cos2x, n), X belongs to R and f (x) = AB, if the image of function f (x) passes through points (0,1) and (Pie / 4,1), M = 1, n = 1 Find the minimum positive period of 1: F (x) and find the minimum value of F (x) on X ∈ [0, Π / 4] 2: When f (A / 2) = 1 / 5 and a belongs to one or two image limits, find the value of sina

1.∏,min=1
2.cosa=-3/5 ,sina=4/5

2A vector = (sin2x, cos2x), B = (cos2x, - cos2x) A vector = (Radix 3sin2x, cos2x), B = (cos2x, - cos2x) (1) X ∈ (7 / 24 Π, 5 / 12 Π), a * B + 1 / 2 = - 3 / 5, find cos4x (2) The three sides of the triangle ABC are a, B, C respectively, and B * b = AC, the angle corresponding to B side is x, and a vector * B vector + 1 / 2 = m has and only one real root. Find M

(1)
a*b+1/2=√3sin2xcos2x-cos^2(2x)+1/2
=(√3/2)sin4x-1/2(cos4x+1)+1/2
=(√3/2)sin4x-(1/2)cosx-1/2+1/2
=-cos(4x+∏/3)
=-3/5
So cos (4x + Π / 3) = 3 / 5
Because x ∈ (7 / 24 Π, 5 / 12 Π)
4x+∏/3∈(3/2∏,2∏)
So sin (4x + Π / 3) < 0
sin(4x+∏/3)=-4/5
So cos (4x) = cos (4x + Π / 3 - Π / 3)
=cos(∏/3)cos(4x+∏/3)+sin(∏/3)sin(4x+∏/3)
=(1/2)*(3/5)+(√3/2)*(-4/5)
=(3-4√3)/10
(2)
cosx=(a^2+c^2-b^2)/2ac
B ^ 2 = AC
cosx=(a^2+c^2-ac)/2ac
≥＜2√(a^2*b^2）-ac＞/2ac
=(2ac-ac)/2ac
=1/2
Then x ∈ [Π / 3,2 Π / 3]
B ^ 2 = AC
Then B is not the largest edge, so x is less than 90
Then x ∈ [Π / 3, Π / 2)
Then 4x + Π / 3 ∈ [5 Π / 3,7 Π / 3)
A vector * B vector + 1 / 2 = - cos (4x + Π / 3)
The period is Π / 2
So - cos (4x + Π / 3) is not monotonic on Π / 3,2 Π / 3]
And a vector * B vector + 1 / 2 = m has and only one real root
So x = Π / 3
m=-cos(4∏/3+∏/3)
=-1/2
To be honest, I'm not sure about the second question
I don't know if I can help,

If x ∈ (0,4 / π), find the value range of the function y = cos2x-sin2x + 2sinxcosx

y=cos2x-sin2x+2sinxcosx
=cos2x-2sinxcosx+2sinxcosx
=cos2x
x∈(0,4/π)
2x∈(0,2/π)
So the range is (0,1)

It is known that 0 ≤ x ≤ π Then the function y = 4 The value range of 2sinxcosx + cos2x is______ .

The original formula can be reduced to y = 3sin (2x + φ), where cos φ = 2
Two
3，sinφ=1
And there are φ≤ 2x + φ ≤ π + φ
∴ymax=3sinπ
2=3，
ymin=3sin（π+φ）=-3sinφ=-1．
The value range is [- 1,3]
So the answer is [- 1,3]

The known function f (x) = (sin2x cos2x + 1) / 2sinx (1) Find the definition domain of F (x) (2) Let α be an acute angle and sin α = 4 / 5, then find the value of F (α)

1)sinx≠0==>x≠kπ
2) By simplifying the original formula, f (x) = 2 (cosx + SiNx)
A is an acute angle, so cosa > 0
So cosa = 3 / 5
So f (a) = 14 / 5

The known function f (x) = 2sinx ^ 2 + sin2x-1 1. Find the minimum positive period of FX and the set of X when FX takes the maximum value? 2. Finding monotone decreasing interval of FX

f(x)=sin2x-(1-)2sinx^2=sin2x-cos2x
=√2sin(2x-π/4)
1. Minimum positive circumference t = 2 π / 2 = π
The maximum value = √ 2 2x - π / 4 = 2K π + π / 2x = k π + 3 π / 8
2. Monotone decreasing interval
2kπ+π/2

The period of y = sin2x + cos2x / TaNx + Cotx Find the period of this function

Y = sin2x + cos2x / TaNx + Cotx = (sin2x + cos2x) / (2 / sin2x) = 1 / 2 (sin2xsin2x + sin2x cos2x) = 1 / 2 [1 / 2 (1-cos4x) + 1 / 2sin4x] period T = 2 π / 4 = π / 2

Find the value range of the function f (x) = cos2x + 1 / 2sinx

f(x)=cos2x+1/2sinx
=1-2sin2^x+1/2sinx
=-2sin2^x+1/2sinx+1
Let t = SiNx, t belong to [- 1,1]
f(x)=-2t^2+1/2t+1
-b/2a=1/8
Finally, according to the image: F (x) = [- 3 / 2,33 / 32]

The value range of the function f (x) = cos2x + 2sinx, X ∈ (π / 2,3 π / 4) is

f（x)=cos2x+2sinx=1-2sin²x+2sinx=-2（sin²x-sinx+1/4）+1/2+1=-2（sinx-1/2）²+3/2x∈(π/2,3π/4)-2（1-1/2）²+3/2＜f（x)＜-2（√2/2-1/2）²+3/21＜f（x)＜√2