It is known that the function f (x) is a decreasing function in the domain of definition (- infinity, 4) and can make f (m-sinx) ≤ f (1 + 2m) - 7 / 4 under the root sign)

It is known that the function f (x) is a decreasing function in the domain of definition (- infinity, 4) and can make f (m-sinx) ≤ f (1 + 2m) - 7 / 4 under the root sign)

Because f (x) is a decreasing function in the domain of definition (- ∞, 4), as long as
4≥√（1+2m） - 7/4 + Cos^2 X≥m-sinx
That's fine
①
From 4 ≥ √ (1 + 2m) - 7 / 4 + cos ^ 2 x →
1+2m≥0;→m≥-1/2;
If cos ^ 2 x ≤ 23 / 4 - √ (1 + 2m), then if cos ^ 2 x ≤ 1 and x = R, there must be
23/4-√（1+2m）≥1.
The solution is m ≤ 357 / 32
M ≥ - 1 / 2,
∴-1/2≤m≤357/32.
②
4 ≥ m-sinx
sinx≥m-4;
If the range of trigonometric function SiNx ≤ 1 and x = R, then there must be
m-4≤1;
→m≤5.
③
From √ (1 + 2m) - 7 / 4 + cos ^ 2 x ≥ m-sinx, it is concluded that: 1
Cos^2 X +sinx ≥m-√（1+2m）+ 7/4
Then - 2Sin ^ 2x + SiNx + 1 ≥ m - √ (1 + 2m) + 7 / 4
-2(sinx -1/4)^2 +9/8 ≥ m-√（1+2m）+ 7/4.
Then from the range of trigonometric function - 1 ≤ SiNx ≤ 1, we can get - 2 ≤ - 2 (SiNx - 1 / 4) ^ 2 + 9 / 8 ≤ 9 / 8
And if x = R, then there must be
-2≤ m-√（1+2m）+ 7/4≤9/8
These are two inequalities; solve them separately
From - 2 ≤ m - √ (1 + 2m) + 7 / 4, it is obtained that: 1
Tick (1 + 2m) ≤ m + 2 → square
1+2m≤ m^2+4m+4;
Then m ^ 2 + 2m + 3 ≥ 0; m ∈ R;
From M - √ (1 + 2m) + 7 / 4 ≤ 9 / 8, it is concluded that: 1
M-5 / 8 ≤ √ (1 + 2m); → square
1+2m≤ m^2-(5/4)m+25/64;
Thus M can be determined
Taking the intersection of ①, ② and ③ is the value range of real number M
//Yes, I calculate it according to the formula under the root sign, which is only (1 + 2m)
Oh, I Baidu search your topic