It is known that the minimum positive period of the function y = 2Sin (Wx - Wu / 3) (where w > 0) is Wu. (1) find the value of W. (2) find the function It is known that the minimum positive period of the function y = 2Sin (Wx - U / 3) (where w > 0) is u (1) Find the value of W （2） Find the maximum value of the function and the corresponding x value when the maximum value is obtained

It is known that the minimum positive period of the function y = 2Sin (Wx - Wu / 3) (where w > 0) is Wu. (1) find the value of W. (2) find the function It is known that the minimum positive period of the function y = 2Sin (Wx - U / 3) (where w > 0) is u (1) Find the value of W （2） Find the maximum value of the function and the corresponding x value when the maximum value is obtained

(1) The minimum positive period of the function y = 2Sin (Wx - π / 3) (where w > 0) is π
So t = 2 π / w = π
We get w = 2
(2) So f (x) = 2Sin (2x - π / 3)
Let 2x - π / 3 = π / 2 + 2K π (k is an integer)
X = k π + 5 π / 12 (k is an integer)
So when x = k π + 5 π / 12 (k is an integer), f (x) takes the maximum value of 2

Calculus ∫?. √ (4-x? 2) DX=

And Y is greater than 0, and X is 0-2, so it is the area of the first quadrant, pi * r * 1 / 4 = Pi

To get the image of the function y = cos (2x - π / 6), we only need to transform the image of function y = cos2x A shifts π / 6 B shifts π / 12 to the right

To get the image of function y = cos (2x - π / 6) = cos [2 (x - π / 12)], we only need to shift the image of function y = cos2x (shift π / 12 units to the right)

How does 1 + sin2x + 2cos ^ 2x-1 + 1 get sin2x + cos2x + 2? I don't understand. If there are detailed steps, it would be good,

2cos^2x-1=cos2x
1+sin2x+2cos^2x-1+1
=sin2x+cos2x+2

How can (SiNx + cosx) ^ 2 + 2cos ^ 2x-2 change to 1 + sin2x + 1 + cos2x-2 Want detailed process ~!

(sinx+cosx)^2+2cos^2x-2
=(sin²x+cos²x+2sinxcosx)+(2cos²x-1)-1
=1+sin2x+cos2x-1
=1+sin2x+1+cos2x-2

SiNx + cosx = 0.2, X belongs to the process of (0, PI) calculating TaNx value!

sinx+cosx=0.2
Both sides square and (SiNx) ^ 2 + (cosx) ^ 2 = 1
So 1+2sinxcosx=0.04
sinxcosx=-0.48
sinx+cosx=0.2
So SiNx and cosx are the solutions of the equation a ^ 2-0.2a-0.48 = 0
(a-0.8)(a+0.6)=0
a=0.8,a=-0.6
X belongs to (0, PI)
So SiNx > 0
So sin = 0.8, cosx = - 0.6
tanx=sinx/cosx=-4/3

Given SiNx + cosx = 1 / 5, X range (0, PI), find TaNx

∵sinx+cosx=1/5
∴(sinx+cosx)²=sin²x+2sinxcosx+cos²x=1/25
∵sin²x+cos²x=1
∴sinxcosx=-12/25
Substituting SiNx = 1 / 5-cosx into
25cos²x-5cosx-12=(5cosx+3)(5cosx-4)=0
/ / cosx = - 3 / 5 or cosx = - 4 / 5
When cosx = - 3 / 5, SiNx = 4 / 5, which satisfies the condition of X range (0, PI)
When cosx = 4 / 5, SiNx = - 3 / 5, which does not satisfy the condition of X range (0, PI)
∴tanx=sinx/cosx=-4/3

If SiNx = 2cosx, then sin2x + 1 = 2___ .

∵sinx=2cosx，∴tanx=2．
Then sin2x + 1 = sin2x
sin2x+cos2x+1=tan2x
tan2x+1+1=22
22+1+1=9
5．
So the answer is: 9
5．

tanX=3,pi

Analysis:
π is known

Arcsin ((SiNx) ^ 2) derivation, The answer secx doesn't come out

arcsin((sinx)^2)'=2sinxcosx/√（1-（sinx）^4）