The value range of 2cosx-1 under y = radical is

The value range of 2cosx-1 under y = radical is

In order to make the function meaningful, 2cosx-1 ≥ 0
cosx≥½
So {x ▏ 2K π - π / 3 ≤ x

2. The value range of (1-2cosx) under the radical of function y = is () for detailed explanation,

answer:
1-2cosx∈[-1,3]
And 1-2cosx ≥ 0
∴ 1-2cosx∈[0,3]
The value range of (1-2cosx) under the function y = radical is [0, √ 3]

Find the value range of the function y = 1 / {radical (2cosx-1)}

-1

F (x) = the range of (1 + SiNx) + 1-sinx

The definition domain is r
y>=0
y^2=1+sinx+1-sinx+2√(1-sinx)(1+sinx)
=2+2√[1-(sinx)^2]
=2+2√(cosx)^2
=2+2|cosx|
0<=|cosx|<=1
0<=2|cosx|<=2
2<=2+2|cosx|<=4
2<=y^2<=4
y>=0
√2<=y<=2
Range [√ 2,2]

The value range of 1-2 SiNx under y = radical

First of all - 1 "SiNx" 1,
So 1 / 2 "1-2 / SiNx" 3 / 2,
So the range of values is a closed interval between two and six

The value range of SiNx cosx under y = radical

This problem skillfully uses the sum of two corners formula
y=√（sinx-cosx）=√【(√2)(cosπ/4sinx-sinπ/4cosx)】=2^(1/4)*√sin(x - π/4)
So the range is [0,2 ^ (1 / 4)]

Find the definition domain and value domain of the following function (1) y = SiNx + 3 / SiNx + 2 (2) y = root sign (1 + SiNx)

(1) The domain of definition is positive and negative infinity
The range is [2 / 3,4]
(2) The domain of definition is positive and negative infinity
The value range is [0, root 2]

Kneel down to seek the master of mathematics ~ ~ ~ get the value range ① y = 3-2sin2x ② y = | SiNx | + SiNx ③ y = 1-1 / 2 SiNx under the root sign, and the steps should be detailed~ And there's more Given the function y = 1 + 2sinx, if x=_____ When, y takes the maximum value____ ; when x=____ When y goes to the minimum____ Given the function y = 1-2sinx, when x=_____ When, y takes the maximum value____ ; when x=____ When y goes to the minimum____ Thank you!

My God, so many questions... You don't want to do this homework
1. Because the value range of sin2x is [- 1,1]
So the range of y = 3-2sin2x is [1,5]
2. Discuss in sections
When SiNx

Simplification (1 + SiNx / 1-sinx) + (1-sinx / 1 + SiNx under radical)

Under the first radical, the numerator and denominator are multiplied by 1 + SiNx, and the numerator and denominator are multiplied by 1-sinx
The result is as follows: 1

Simplify the following: ①cos(α−π 2) sin(5π 2+α)•sin(α−2π)•cos(π−α)； II. 1+sinα 1−sinα− 1−sinα 1 + sin α, where α is the second quadrant angle

① The original formula = sin α cos α· sin α · (- cos α) = - sin2 α; ②