If 3 sin2x + cos2x = K + 1 has two distinct real roots in [0, π / 2], then the value range of K is There's a radical in front of it. 3. I can't type it

√ 3 sin2x + cos2x in one deformation
=2sin(2x+π/6)
Draw the image of 2Sin (2x + π / 6) on [0, π / 2]
Draw a line parallel to the x-axis, move up and down, and find out the range of y when there are two intersections between the line and the 2Sin (2x + π / 6) image, which is the range of K + 1
The range of K is then solved
Points good little, if the answer is right, give a good comment!

It is known that the equation cos2x - √ 3 sin2x = k, (x belongs to [- π / 6, π]) has three real solutions, and the value range of K is obtained

cos2x-√3 sin2x=k
2cos(2x+π/3)=k
-π/6≤x≤π
0≤2x+π/3≤7π/3
Draw images
As can be seen from the figure,
1≤k

If y = sin2x and y = cos2x are mostly increasing functions, find the value range of X It's better to write down the process. Thank you

0

(sinX)∨2＝2(cosx)∨2-1

How much is 1 + SiNx and 1 + cosx, like a formula or something?

1+sinx=(sin(x/2)+cos(x/2))^2
1+cosx=2cos(x/2)^2

SiNx and cosx conversion What are there?

sin²x+cos²x=1

What is the formula of the required 4 descending power in senior one mathematics

Only when the power is reduced and the angle is raised can (cosx) ^ 2 = (1 + cos2x) / 2

On the formula of decreasing power in mathematical problems of senior one sin²x=(1-cos2x)/2 cos²x=（1+cos2x）/2 How these two formulas are derived, please tell me more details, thank you

cos2x=1-2sin²x==>sin²x=(1-cos2x)/2
cos2x=2cos²x-1==>cos²x=(1+cos2x)/2

Proof: 2 (COS α - sin α) 1+sinα+cosα＝cosα 1+sinα−sinα 1+cosα．

Method 1: right = cos α + Cos2 α - sin α - Sin 2 α (1 + sin α) (1 + cos α) = (COS α - sin α) (1 + cos α + sin α) 1 + sin α · cos α + sin α = 2 (COS α - sin α) (1 + cos α + sin α) 2 (1 + sin α + cos α + sin α) = 2 (COS α - sin α) (1 + cos α + sin α)

If 3 sin2x + cos2x = K + 1 has two distinct real roots in [0, π / 2], then the value range of K is There's a radical in front of it. 3. I can't type it