Sin ^ 3x cos ^ 3x ≥ cosx SiNx, find the value range of X. x ∈ {0,2 π}

Sin ^ 3x cos ^ 3x ≥ cosx SiNx, find the value range of X. x ∈ {0,2 π}

sin^3x-cos^3x≥cosx-sinx
(sinx-cosx)(sin^2x+sinxcosx+cos^2x)+(sinx-cosx)≥0
(sinx-cosx)(1+sinxcosx+1)≥0
2 + sinxcosx constant > 0
∴sinx-cosx≥0
sin≥cosx
∵x∈（0,2π）
∴x∈(π/4,5π/4)

If SiNx + cosx = root 2, find the value of sin ^ 3x + cos ^ 3x

sin^3X+cos^3X
=（sinx+cosx)(sin^2x-sinxcosx+cos^2x)
=Radical 2 (1 + sinxcosx)
SiNx + cosx = radical 2
So sin ^ 2x + cos ^ 2x + 2sinxcosx = 2, so sinxcosx = 1 / 2
So the original formula is equal to (radical 2) / 2

If the image of the function y = cosx is shifted to the left by a unit (0 ≤ a ≤ 2 π), the image of the function y = cos (x - π / 6) is obtained, then a =?

The image of function y = cosx shifts π / 6 units to the right and becomes the image of function y = cos (x - π / 6)
(left plus right minus)
Because the period of the function is 2 π
So it can be said that the left shift is 2 π - π / 6 = 11 π / 6
So a = 11 π / 6

The function y = cos (3x + π) 3) The image of y = cosx can be changed from the image of y = cosx to______ Translation______ The abscissa of all the points on the resulting image are then taken______ For the original______ It is obtained by multiplying (the ordinate remains unchanged)

The image with y = cosx shifts π to the left first
Three units,
Then change the ordinate of each point to the original 1
Three times,
Then y = cos (3x + π) can be obtained
3) The image of
So the answer is: left; π
3. Shrink; 1
3．

The image of the function y = f (x) × cosx is translated by vector a = (PAI / 4,1) to obtain the image of function y = 2sinx ^ 2, then the function f (x) can be What does a cosx B 2sinx C SiNx D 2cosx and vector translation mean?

The image with y = f (x) × cosx is translated by vector a = (PAI / 4,1)
That is, shift π / 4 units to the right and 1 unit up
The analytic formula is y = f (x - π / 4) cos (x - π / 4) + 1, which is reduced to y = 2Sin? X
If a cosx, then y = cos 2 (x - π / 4) + 1 cannot be converted into 2Sin 2 X
If B 2sinx, then y = 2Sin (x - π / 4) cos (x - π / 4) + 1 = 1 + sin (2x - π / 2) = 1-cos2x = 2Sin? X (correct)
If C SiNx, then y = sin (x - π / 4) cos (x - π / 4) + 1 = 1-1 / 2cos2x cannot be converted into 2Sin? X
If D 2cosx, then y = 2cos 2 (x - π / 4) + 1 cannot be converted into 2Sin? X
Choose B

The image of the function y = f (x) cosx is translated by vector a = (PAI, 1) to obtain the image of y = 2Sin ^ 2x

Translate by vector a = (π / 4,1), first shift π / 4 units to the right, and then 1 unit up
Y = 2Sin ^ 2x = 1-cos2x
First shift π / 4 units to the left and then 1 unit down:
y+1=1-cos2(x+π/4)
y=-cos(2x+π/2)=sin2x=2sinxcosx
∴f(x)=2sinx

The periodic and monotone intervals of the function F X = sin ^ 2x + √ 3sinxcosx + cos ^ 2 are known

Double angle formula: cos2x = cos? X-sin? 2x = 1-2sin? X, so sin? X = (1-cos2x) / 2sin2x = 2sinxcosx, so sinxcosx = (sin2x) / 2Sin ^ 2x + √ 3sinxcosx = (1-cos2x) / 2 + √ 3 (sin2x) / 2 = √ 3 / 2 × sin2x-1 / 2 × cos2x-1 / 2 (√ 3 / 2 and

Let f (x) = sin ^ 2 + 2 √ 3sinxcosx cos ^ 2x (1) find the minimum positive period of F (x) and (2) find the maximum of F (x) on the interval [0, π / 2] (1) Find the minimum positive period of function f (x) (2) find the maximum and minimum value of F (x) on the interval [0, π / 2]

According to the formula asin (Wx + n). T = 2 π / W, so the minimum positive period is t = 2 π / w = 2 π / 2 = π 2. Find the maximum and minimum value. From the 2Sin (2x - π / 6) of 1, we can know that the positive selection function

Finding the minimum positive period of F (x) by known function f (x) = cos 2 x / 2-sin 2 x / 2 + sin + When x0 ∈ (0, π / 4) and f (x0) = 5 / 2 of the root sign of 4, find the value of F (x0 + π / 6)

F (x) = cos? X / 2 - sin? X / 2 + SiNx = cosx + SiNx = root 2 (sinxcos π / 4 + cosxsin π / 4) = Radix 2 sin (x + π / 4) has the minimum positive period 2 π x0 ∈ (0, π / 4) and f (x0) = 5 / 2 of radical 2, namely: root 2 sin (x0 + π / 4) = 5 / (4 radical 2) sin (x0 + π / 4) = 5 / 8

The vector a = (sin (2x + θ), cos (2x + θ), B = (1, radical 3), function f (x) = AB is even function, and θ∈ [0, π] 2. Let x ∈ (0, π / 2), f (x) = 1, find the value of X

Solution by F (x)
=a*b=sin（2x+θ）+√3cos（2x+θ）
=2[1/2sin（2x+θ）+√3/2cos（2x+θ）]
=2sin（2x+θ+π/3)
Let f (x) = AB be an even function and θ∈ [0, π]
Then θ + π / 3 = π / 2
That is, θ = π / 6
So f (x) = 2Sin (2x + π / 6 + π / 3)
=2sin（2x+π/2)
=2cos2x
2 by X ∈ (0, π / 2)
Then 2x ∈ (0, π)
By F (x) = 2cos2x = 1
That is, cos2x = 1 / 2
So 2x = π / 3
That is, x = π / 6