Find the value range of 1-2sinx under the function y = 3sinx + radical No, I'm sorry

y=3sinx+√(1-2sinx) -1

The value range of function y = | cosx | + 3cosx

Discuss removing absolute values
When x ∈ [- π / 2, π / 2], cosx > = 0, y = 4cosx
When x ∈ [π / 2,3 π / 2], cosx

The value range of function y = cosx / 2-3cosx

Y = cosx / 2-3 * 2cos? X / 2 + 3 = - 6 (COS? X / 2-1 / 6cosx / 2) + 3 = - 6 (cosx / 2-1 / 12) 2 + 3 + 1 / 24 = - 6 (cosx / 2-1 / 12) 2 + 73 / 24 when cosx / 2 = 1 / 12, y gets the maximum value: ymax = 73 / 24. When cosx / 2 = - 1, y gets the minimum value: Ymin = - 6 (- 1-1 / 12) mm2

Find the value range of the following function:) y = (cosx) ^ 2-3cosx + 1 (x ∈ [- π / 4, π / 3]) )y=(cosx)^2-3cosx+1 （x∈[-π/4,π/3])

Let cosx = t, because x ∈ [- π / 4, π / 3], t = cosx ∈ [1 / 2,1]
Y = T ^ 2-3T + 1 derivative y '= 2t-3 is always less than 0, that is, y monotonically decreases
So, when t = 1 / 2, y max = - 1 / 4
When t = 1, the minimum value of y = - 1
Therefore, its value range is [- 1, - 1 / 4]
If you haven't learned the derivative, you can also make a formula for y = T ^ 2-3T + 1, and draw a graph to know its maximum and minimum points

Find the range of y = (3cosx + 1) / (cosx + 2)

The function y = (3cosx + 1) / (cosx + 2)] can be used to calculate the range of the whole formula. Y = (3cosx + 1) / (cosx + 2) = [3 (cosx + 2) - 5] / (cosx + 2) = 3-5 / (cosx + 2) ∵ cosx ∈ [- 1,1],  cosx + 2 ∈ [1,3],  5 / (cosx + 2) M cosx ∈ [- 1,1],  cosx + 2 ∈ [1,3],  - 5 / (cosx + 2

Given a = (2sinx, m), B = (SiNx + cosx, 1), the function f (x) = AB if the maximum value of F (x) is the root sign 2. Find the value of M (2) If f (x) is shifted to the left by n (n > 0) units and symmetric about the y-axis, find the minimum value of n

1. F (x) = AB = 2 (SiNx) ^ 2 + sin2x + M = 1-cos2x + sin2x + M = (radical 2) * sin (2x pi / 4) + m + 1
So the maximum value of F (x) = (root 2) + m + 1 = root 2
So m = - 1
2. From (1), the symmetry axis nearest to the X axis is x = 3pi / 8,
If so is symmetric about the y-axis, then n = 3pi / 8

Let a = (2sinx, m), B = (SiNx = cosx, 1), f (x) = AB (x ∈ R), if the maximum value of F (x) is root two 1. Find the value of M 2. If the image of F (x) is shifted to the left by every unit n (n > 0), the minimum value of n is obtained

F (x) = 2sinx ^ 2 + m; since SiNx = cosx, then 2Sin ^ 2 = 1; therefore, M = radical 2-1;
Because SiNx = cosx, then x = π / 4 + K π; because of symmetry about y-axis, x = π / 4 when k = 0; therefore, n = π / 4;

Let a = (2sinx, m), B = (SiNx + cosx, 1), and the function f (x) = AB if the maximum value of F (x) is the radical 2 (2) If f (x) is shifted to the left by n (n > 0) units and symmetric about the y-axis, find the minimum value of n

F (x) = 2 (SiNx) ^ 2 + 2sinxcosx + M = 1-cos (2x) + sin (2x) + M = 2 ^ 0.5 * sin (2x pi / 4) + (M + 1) f (x) max root 2, so m = - 1, f (x) = 2 ^ 0.5 * sin [2 (x-pi / 8)] shift pi / 8 to the left to get f (x) = 2 ^ 0.5 * sin (2x) and then shift pi / 4 to the left to get f (x) = 2 ^ 0.5 *

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

Change the element, change the root sign (1-2sinx) into t, write out the range of T, and then change the variable into a quadratic function, and then solve the value range according to the maximum and minimum value of the quadratic function in the definition domain, that is, the value range of the original function

SiNx + root sign 3cosx + a = 0 has two different real roots on X ∈ [0, π / 2]

3cosx + a = 0 under SiNx + radical
3cosx = 2Sin (x + π / 3) under SiNx + radical
x∈[0,π/2] x+π/3∈[π/3,5π/6]
2sin(x+π/3)∈[1,2]
Because there are two radicals, x + π / 3 ∈ [π / 3,2 π / 3]
2Sin (x + π / 3) ∈ [3,2 under radical]
So a ∈ [- 2, - 3] under the radical
That is - 2

Find the value range of 1-2sinx under the function y = 3sinx + radical No, I'm sorry