If the function f (x) = SiNx + | SiNx |, X ∈ [0,2 π] has only two different intersections with the line y = k, then the value range of K is

If the function f (x) = SiNx + | SiNx |, X ∈ [0,2 π] has only two different intersections with the line y = k, then the value range of K is

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When x belongs to [0, π], f (x) = 2Sin X; when x belongs to [π, 2 π], f (x) = 0
If you draw an image, y = k has two intersections, so k = (0,2). If you take 2, there will be one, if you take 0, there will be countless

Given the function f (x) = 3sin2x-2sin2x. (I) find the maximum value of function f (x); (II) find the set of zeros of function f (x)

(I) ∵ f (x) = 3sin2x-2sin2x = 3sin2x + cos2x-1 = 2Sin (2x + π 6) - 1, so the maximum value of function f (x) is equal to 2-1 = 1 (II) from F (x) = 0 we get 23sin xcos x = 2sin2x, then SiN x = 0, or 3cos x = SiN x, that is, Tan x = 3, from sin x = 0 we know x = k π; from Tan x = 3 we know x = K

Given the function f (x) = 3sin2x-2sin2x. (I) find the maximum value of function f (x); (II) find the set of zeros of function f (x)

(I) ∵ f (x) = 3sin2x-2sin2x = 3sin2x + cos2x-1 = 2Sin (2x + π 6) - 1, so the maximum value of function f (x) is equal to 2-1 = 1 (II) from F (x) = 0 we get 23sin xcos x = 2sin2x, then SiN x = 0, or 3cos x = SiN x, that is, Tan x = 3, from sin x = 0 we know x = k π; from Tan x = 3 we know x = K

The maximum value of the function y = 3sin & # 178; x-3sin2x + 11cos & # 178; X is

The answer is 4
=3sin²x+3cos²x-3sin2x+8cos²x
=8(cos2x-1)/2-3sin2x+3
=4cos2x-3sin2x-1
The maximum value is 4 and the minimum value is - 6