It is known that the equation cos & # 178; X + 4sinx-a = 0 has a solution, and the value range of a is cos & # 178; X + 4sinx-a = 1-sin & # 178; X + 4sinx-a Why can't we use B ^ 2-4ac ≥ 0 The solution is 4 ^ 2 - (a + 1) ≥ 0

It is known that the equation cos & # 178; X + 4sinx-a = 0 has a solution, and the value range of a is cos & # 178; X + 4sinx-a = 1-sin & # 178; X + 4sinx-a Why can't we use B ^ 2-4ac ≥ 0 The solution is 4 ^ 2 - (a + 1) ≥ 0

Firstly, SiNx is replaced by T, and the equation becomes 1-T ^ 2 + 4t-a = 0;
But please note that t has a value range [- 1,1]
If the equation has a solution, it means that the solution is on [- 1,1],
And the discriminant method you said is for t in all real numbers
The solution of this problem is as follows:
First, change to the form of 1-T ^ 2 + 4t-a = 0, and then change to a = 1-T ^ 2 + 4T,
The value range of a is the value of the right quadratic function on [- 1,1]
The result is that the right formula is 5 - (T-2) ^ 2, and the value range is [- 4,4]
So the value range of a is [- 4,4]

When can we use the root discriminant method to find out whether the equation has a real root? For example, we know that the equation cos & # 178; X + 4sinx-a = 0 has a solution and the value range of A
In this question, why can't we use the root discriminant method to find the range of values

In algebra, we can use the discriminant of roots to find out whether the equation has real roots

If the equation cos & # 178; X + 4sinx + C = 0 has a solution in [0, π], what is the range of C

sinx+cosx=k
k=√2sin（x+π/4）
0≤x≤π
So π / 4 ≤ x + π / 4 ≤ 5 π / 4
So - √ 2 / 2 ≤ sin (x + π / 4) ≤ 1
-1≤√2sin（x+π/4）≤√2
So the value range of K is [- 1, √ 2]

If the equation 2cos2x-4sinx + 4K + 5 = 0 about X has a solution, then the value range of the real number k is______ ．

The original equation can be reduced to k = (SiNx + 12) 2 − 2, ∵ - 1 ≤ SiNx ≤ 1 ∵ - 2 ≤ (SiNx + 12) 2 − 2 ≤ 14