Find the value range of function f (x) = 2 (SiNx + cosx) + sin2x-1

Find the value range of function f (x) = 2 (SiNx + cosx) + sin2x-1

f(x)=2(sinx+cosx)+sin2x-1 =2√(sin2x+1)+(sin2x+1)-2 =[√(sin2x+1)+1]²-3∵0≤sin2x+1≤2∴0≤√(sin2x+1)≤√2∴1≤√(sin2x+1)+1≤√2+1∴1≤[√(sin2x+1)+1]²≤3+2√2-2≤[√(sin2x+1...

Given the vector M = (2sinx cos, SiNx) vector n = (cosx SiNx, 0) and f (x) = (M + 2n) m (1), find the minimum positive period of function f (x) (2) By shifting the function f (x) to the left π / 4 units, the monotone increasing interval of G (x) is obtained

(M + 2n) = (cosx, SiNx), (M + 2n) * m = (2sinx-cosx) cosx + SiNx = 2ssinx cosx-cos-s (x + SiNx) x + sin = sin2x-cos2x = sin2x-cos2x = sin2x-cos2x = sin2x (2x-π / 4) therefore, f (x) = √ 2Sin (2x - π / 4) minimum positive cycle T = π g (x = f (x + π / 4) minimum positive cycle T = π g (x = f (x + π / 4) - f (x + π / 4) - π / 4) = √ 2Sin [2 (x + π / 4) - π / 4] = √ 2Sin (2x (2x (2x / plus

Given the vector m (cosx, - SiNx), vector n (cosx, sinx-2, radical 3cosx), X belongs to R, Let f (x) = m * n + 2, Given vector m (cosx, - SiNx), vector n (cosx, sinx-2 root sign 3cosx), X belongs to R, Let f (x) = m * n + 2, find the minimum value of function f (x); 2. If f (x) = 50 / 13, and X belongs to [π / 4, π / 2], find the value of sin2x. 3. Let the inequality f (x) > = 3 hold. 4. If the equation f (x) = K (0)

Problem 1: so f (x) = (cosx) ^ 2 - (SiNx) ^ 2 + 2 radical sign 3sinxcosx + 2
=(cos2x + 1) / 2 - (1-cos2x) / 2 + radical 3sin2x
=Cos2x + Radix 3sin2x
=2 * (1 / 2cos2x + radical 3 / 2sin2x)
=2 * (sin30 degree cos2x + cos30 degree sin2x)
=2 * sin (2x + 30 degrees)
Because x belongs to R, there is a minimum value of 0 when x = 5 π / 12 + K π
There's another question I don't have time to answer, but it seems to be a senior one or from last semester?

If x ∈ [π / 6, π / 3], then the value range of the function f (x) = 2cos? X + sinx-1 is?

0-1

The value range of function f (x) = 2cos ^ x + sinx-1, X ∈ [0, π / 2]

Substitution method
Let SiNx = t,
∵ x∈[0,π/2]
∴ t∈[0,1]
∴ f(x)=2cos²x+sinx-1
=2-2sin²x+sinx-1
=-2sin²x+sinx+1
=-2t²+t+1
=-2(t-1/4)²+9/8
When t = 1 / 4, f (x) has a maximum value of 9 / 8
When t = 1, f (x) has a minimum value of 0
The value range of F (x) is [0,9 / 8]

Given the function f (x) = a (2cos ^ 2x / 2 + SiNx) + B, when a = 1, find the monotone increasing interval of function f (x);

B = a (1 + cosx + SiNx) + B = a (1 + cosx + SiNx) + B = a (1 + cosx + SiNx) + B = a (cosx + SiNx) + A + B = √ 2A (sin π / 4cosx + cos π π / 4sinx) + A + B = √ 2asin (π / 4 + x) + A + B + B + B + monotone increasing interval is [2K π + 3 π / 2-π / 4, (2k + 1) π / 4, (2k + 1) π / 4, (2k + 1) π + π / 2-π / 4] that is: [2K π + 5 π / 4, (4, (2k π / 4), (2k π / 4), (2k π + 5 π / 2K + 1) π + π / 4]

Let f (x) = SiNx radical 3cosx, X ∈ R (1) find the minimum positive period of F (x) and (2) find the monotone increasing interval of F (x)

f(x)=sinx-√3cosx=2sin(x-π/3)
So period T = 2 π
From 2K π - π / 2

Find the minimum value of the function y = SiNx + 2sinx · cosx + 3cosx and the set of X with the minimum value, and find its maximum value

(2x + π / 4) when 2x + π / 4 = 2K π + π / 2) when 2x + π / 4 = 2K π + π / 2, there is a maximum value of 2 + √ 2, x = k π + π / 8, K ∈ Z, when 2x + π / 4 = 2K π + π / 2, K ∈ Z, when 2x + π / 4 = 2K π - π / 2, K ∈ Z, when 2x + π / 4 = 2K π - π / 2, K ∈ Z, when 2x + π / 4 = 2K π - π / 2, X = k ∈ 2, x = k ∈ 2, x = k ∈ 2, x = k ∈ 3 Z

Find the tangent equation of the curve X ^ 2 / 3 + y ^ 2 / 3 = a ^ 2 / 3 at the point ((root 2) / 4 A, (root 2) / 4A)

The original equation is a circle equation, (root 2) / 4A, (root 2) / 4A) is a point on the circle, so the tangent passing through this point is perpendicular to the straight line from the center of the circle to this point, and the center of the circle is (0,0)
So the first line is y = x, the slope of the line is 1, the slope of the tangent is - 1, so the tangent equation is: y = - x + √ 2 / 2A

Find the tangent equation and normal equation of X at point m (1,1) under the curve y = f (x) = cubic root

y=x^1/3
y`=1/3x^(-2/3)
k=y`(x=1)=1/3
y-1=1/3(x-1)
y-1=-3(x-1)