Find the value range of function y = (2-cosx) / SiNx

Find the value range of function y = (2-cosx) / SiNx

Value range: (- infinity, -√ 3] u [√ 3, + infinity) analysis: y = (2-cosx) / SiNx, that is, ysinx = 2-cosx, so ysinx + cosx = 2. According to the formula asinx + bcosx = [√ (a ^ 2 + B ^ 2)] sin (x + T), where tant = B / A, so √ (y ^ 2 + 1) sin (x + T) = 2, so let m = sin (x + T) = 2 / √ (y ^ 2 + 1). Obviously

It is known that (2Sin ^ 2x + sin2x) / (1 + TaNx) = 1 / 2 (π / 4

（2sin^2x+sin2x）/(1+tanx)
=cosx*（2sin^2x+2cosx*sinx）/(1+tanx)*cosx
=2sinx*cosx(sinx+cosx)/(cosx+sinx)
=2sinx*cosx=sin2x=1/2
(sinx-cosx)^2=sin^2x+cos^2x+2sinxcosx
=1+sin2x=3/2
π/40
sinx-cosx=√(3/2)

It is known that cos (π / 4 + x) = 3 / 5,17 π / 12

cosπ/4cosx-sinπ/4sinx=3/5
cosx-sinx=3√2/5
square
cos ² x+sin ² x-2sinxcosx=18/25
1-2sinxcosx=18/25
sinxcosx=7/50
sinx+cosx=√2sin(x+π/4)
5π/4

It is known that - 90 degrees is less than X and less than 0 degrees, SiNx + cosx = 0.2, find the value of (sin2x + 2Sin ^ 2x) / (1-tanx)

Because sin ^ 2x + cos ^ 2x = 1 and SiNx + cosx = 0.2
So 2sinx * cosx = - 0.96 = sin2x
So (SiNx cosx) ^ 2 = 1.96
And because of - 90

It is known that cosx SiNx = three fifths root 2. Find the value of sin2x-2sin ^ 2x / 1-tanx. Please write down the process, O (∩ ∩) O thank you

cosx-sinx=3√2/5cos ² x+sin ² x-2sinxcosx=18/251-sin2x=18/25sin2x=7/25(sin2x-2sin ² x)/(1-tanx)=(sin2x-2sin ² x)/(1-sinx/cosx)=(sin2xcosx-2sinxcosxsinx)/(cosx-sinx)=sin2x(cosx-sinx)/(cosx-...

It is known that cosx SiNx = (3 / 5) * root 2 is afraid of love

Cosx SiNx = (3 / 5) * root 2, square, 1-2sinx * cosx = 18 / 25
2sinx * cosx = 7 / 25, 1 + 2sinx * cosx = 32 / 25, i.e. (SiNx + cosx) ^ 2 = 32 / 25
Because I'm afraid of love