Y = the range of | SiNx | - 2sinx? When SiNx ＞ = 0, y = ｜ SiNx ｜ - 2sinx = - SiNx, the range is [- 1,1] When SiNx < 0, y = - SiNx - 2sinx = - 3sinx Why is the value range of the final answer: [- 1,3]?

Y = the range of | SiNx | - 2sinx? When SiNx ＞ = 0, y = ｜ SiNx ｜ - 2sinx = - SiNx, the range is [- 1,1] When SiNx < 0, y = - SiNx - 2sinx = - 3sinx Why is the value range of the final answer: [- 1,3]?

It is true that y = - SiNx can be obtained from SiNx > = 0, but since it is under the condition of SiNx > = 0, - SiNx should belong to [- 1,0]
Similarly, when SiNx < 0, y = - 3sinx, - 3sinx belongs to (0,3]
The value range of the two is: [- 1,3]. That's right~

The function y = 2sinx + 1 The value range of SiNx − 2 is______ .

∵ function y = 2sinx + 1
sinx−2，
∴y＝2+5
sinx−2．
∵-1≤sinx≤1，
∴-3≤sinx-2≤-1，
∴−1≤1
sinx−2≤−1
3，
∴−5≤5
sinx−2≤−5
3，
∴−3≤y≤1
3，
The original value field of − is 1
3]．
So the answer is [− 3, 1
3]．

Find the range of y = (1-2sinx) / (SiNx + 2) To process

SiNx + 2 ≠ 0, so SiNx ≠ - 2
Because SiNx belongs to [- 1,1]
So x belongs to R
When SiNx = - 1, the maximum value is 3
When SiNx = 1, the minimum value = - 1 / 3
Therefore, the range of the original formula belongs to [- 1 / 3,3]

The steps of (siny) / (siny) / (2)

y=(2sinx+4-5)/(sinx+2)
=(2sinx+4)/(sinx+1)-5/(sinx+2)
=2-5/(sinx+2)
-1

The function y = SiNx (SiNx + cosx) x belongs to the (0, π / 2) evaluation domain

y=sinx(sinx+cosx)
=[1-cos(2x)+sin(2x)]/2
=1/2+√2*[sin(2x)*√2/2-√2/2*cos(2x)]/2
=1/2+√2/2*sin(2x-π/4)
Zero

F (x) = 2sinx (SiNx + cosx). X belongs to R. find 1, the minimum positive period, 2 the maximum value, and find the corresponding set of X, 3 find the monotone decreasing interval of the function

y=2sinx(sinx+cosx)
=2sinxsinx+2sinxcosx
=-（1-2sinxsinx-1+sin2x)
=-(cos2x+sin2x-1)
=-Root 2 ((radical 2) divide 2cos2x + (Radix 2) divide 2sin2x) + 1
=-Radical 2Sin (2x + Π / 4) + 1
Minimum positive period T = Π
Because x ∈ R
So - 1=

What is the range of F (x) = SiNx / 2-cosx?

y=sinx/(2－cosx)
2y－ycosx=sinx
sinx＋ycosx=2y
If [√ (y 2 + 1)] sin (x + W) = 2Y, then:
sin(x＋w)=[2y]/[√(y²＋1)]
Because | sin (x + W) | ≤ 1, then:
|[2Y / √ (y 2 + 1)] | ≤ 1, the square of both sides is obtained
y²＋1≥4y²
If y ≤ 1 / 3, then:
- √ 3 / 3 ≤ y ≤√ 3 / 3, then:
y∈[－√3/3,√3/3]

Find the value range of F (x) = (2-sinx) / (2 + cosx). The answer is [(4 - √ 7) / 3, (4 + √ 7) / 3] For detailed explanation

y=(2-sinx)/(2+cosx)
2+cosx>0
(2+cosx)y=2-sinx
2y+ycosx=2-sinx
sinx+ycosx=2-2y
(1 + y ^ 2) sin (x + φ) = 2-2y,
sin(x+φ)=(2-2y)/√(1+y^2)
-1

Find the value range of the function y = (1-sinx) / (sin2x times cosx)

Y = (1-sinx) / (sin2x * cosx) = (1-sinx) / [2sinx * (1-sinx * SiNx)] = = 1 / [2sinx * (1 + SiNx)] = 1 / 2 [(SiNx + 1 / 2) ^ 2 + 3 / 4] (SiNx + 1 / 2) ^ 2 + 3 / 4] (SiNx + 1 / 2) ^ 2 + 3 / 4] (SiNx + 1 / 2) ^ 2 + 3 / 4, so the range of Y is [1 / 6,2 / 3]

The value range of function f (x) = (sin2x * cosx) / (1-sinx)?

Let t = SiNx
Then f (x) = 2sinx (cosx) ^ 2 / (1-sinx) = 2T (1-T ^ 2) / (1-T) = 2T (1 + T) = (T + 1 / 2) ^ 2-1 / 2
Because|