Given that y = 1, y = x, y = x ^ 2 are three solutions of a second order non-homogeneous linear differential equation, then the general solution of the equation is?

Given that y = 1, y = x, y = x ^ 2 are three solutions of a second order non-homogeneous linear differential equation, then the general solution of the equation is?

If Tan θ > 1 and sin θ + cos θ < 0, then the value range of cos θ is ()
A. (−22，  0)B. (−1，  −22)C. (0，  22)D. (22，  1)

If ∵ Tan θ > 1, the terminal edge of ∵ θ is in the first or third quadrant, and sin θ + cos θ < 0, the terminal edge of ∵ θ is in the third quadrant, then 2K π + 5 π 4 < x < 2K π + 3 π 2, K ∈ Z ∵ 22 < cos θ < 0, so a

It is known that Tan Φ > 1 and sin Φ + cos Φ

Tan Φ > 1 indicates that sin Φ and cos Φ are both positive and negative
sinΦ+cosΦ

If sin α + cos α = Tan α (0

Sin α + cos α = the square of both sides of Tan α = > 1 + 2Sin α cos α = (Tan α) ^ 2 multiple angle formula = > 1 + sin (2 α) = (Tan α) ^ 2 substitute into universal formula = > 1 + [2tan (α)] / {1 + [Tan (α)] ^ 2} = (Tan α) ^ 2 let Tan α = x, 01 + (2x) / (1 + x ^ 2) = x ^ 2 (1 + x) ^ 2 = x ^ 41 + x = x ^ 2x = (1 + radical 5) / 2

If sin θ cubic = cos θ cubic ≥ cos θ - sin θ, and θ ∈ [0,2 π], then the value range of angle θ is?

There are two values
One is π / 4
The other is 5 π / 4

The value range of COS and sin is 111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111

Real number

Let a ∈ (0,2) and point P (sin α, Cos2 α) be in the third quadrant, then the angle α is in the range
Combined with cos2x < 0, we can know that 5 π / 2 < 2x < 7 π / 2

Because P is in the third quadrant, its abscissa and ordinate are negative,
That's Sina

Let θ be the angle of the third quadrant, sin (θ / 2 + 3 π / 2) > 0, then the value of [√ (1-sin θ)] / [cos (θ / 2) - sin (θ / 2)] is?

sin(θ/2+3π/2)>0
-cos(θ/2)>0
cos(θ/2)

θ belongs to the second quadrant angle, sin θ + cos θ =

It should be [- 1,1]. Use the image to solve, draw the image of SiNx and cosx in the rectangular coordinate system at the same time. In the interval, because there is a positive, or a negative, or a positive and a negative, you can get the answer

If cos α = sin α, then the range of the terminal edge of angle α in the first quadrant is ()
A、（0,π/4]
B、[π/4,π/2）
C、[2kπ+π/4,2kπ+π/2],k∈Z
D、（2kπ,2kπ+π/4],k∈Z
Detailed process, thank you

Wrong title:
A falls in the first quadrant
When Sina > cosa, then a ∈ [2K π + π / 4, π / 2]
If Sina = cosa, then a ∈ (2k π, 2K π + π / 4]
When Sina