The function y = cosx The value range of 2cosx + 1 is______ .

Y = cosx
2cosx+1=1
2-1
Two
2cosx+1
∵-1≤cosx≤1，∴-1≤2cosx+1≤3，∴1
Two
2cosx+1≥1
6 or 1
Two
2cosx+1≤−1
Two
The function y = cosx
The value range of 2cosx + 1 is (− ∞, 1
3]∪[1，+∞)
So the answer is (− ∞, 1
3]∪[1，+∞)

The function y = cosx The value range of 2cosx + 1 is______ .

The value range of the function y = cosx / (2cosx + 1) is

Hello, y = cosx / (2cosx + 1) = (cosx + (1 / 2) - (1 / 2)) / (2cosx + 1) = (1 / 2) - (1 / (4cosx + 2)) - 1 / (4cosx + 2) has no maximum or minimum value, because 4cosx + 2 can approach 0, but it is certain that - 1 / (4cosx + 2) cannot be equal to 0, because the denominator is not 0, so the range of Y is (-...)

(cosy = - cosy = - S2) If the title, the process, thank you

y=-2cosx-1
-1

What is the value range of the function y = | cosx | - 2cosx?

Let t = cosx, then | t | = 0, y = t-2t = - t, the range is [- 1,0]
When t

Find the value range of function y = SiNx + 2cosx + 2

y=√5(1/√5sinx+2/√5cosx)+2
=√5(sinx+α)+2
So the value range of the function y = SiNx + 2cosx + 2 is [2 - √ 5,2 + √ 5]

The value range of function y = (SiNx) ^ 2-2cosx

y=(sinx)^2-2cosx=1-(cosx)^2-2cosx=-[(cosx)^2+2cosx+1]+2
=-(cosx+1)^2+2
Because - 1

What is the range of the function y = (2-2cosx) / (sinx-4)?

y=(2-2cosx)/(sinx-4)
ysinx-4y=2-2cosx
ysinx+2cosx=4y+2
√(y^2+4) * [ y/√(y^2+4) *sinx +2/√(y^2+4) *cosx ]=4y+2
Let Sint = 2 / √ (y ^ 2 + 4) cost = Y / √ (y ^ 2 + 4)
√(y^2+4) * sin(x+t)=4y+2
-√(y^2+4)

The value range of function y = 2cos (π / 3) + 2cosx

2cos(π/3)=1
2cosx∈ [-2,2]
therefore
y=2cos(π/3)+2cosx
∈ [-1,3]

Find the value range of function y = (2sinx-1) / (2cosx + 3)

First set cosx to x, the range is - 1 to 1, and then according to SiNx square + cosx square = 1
Find out how many cosx SiNx is equal to, and then bring the SiNx equation about X to the original, and use the formula

The function y = cosx The value range of 2cosx + 1 is______ .