Known sinx 2cosx 3 2sinx 2cosx 0, then the value of sin2x 2cosx 1 tanx is ? Ask God for help

From [sinx-2cosx][3+2sinx+2cosx]=0, sinx-2cosx=0 or sinx+cosx=-3 2 can be obtained because the minimum value of (sinx+cosx) is -root number 2>-3 2, so sinx+cosx=-3 2 is rounded off i.e. sinx-2cosx=0 so sinx=2cosx so sinx cosx=tanx=2 so 1=(sinx) 2+(cosx) 2=(2cosx) 2+(cosx) 2=5(cosx) 2, so (cosx) 2=1 5 so sin2x=2(sinx)*(cosx)=2(cosx)*(cosx) =4(cosx) 2=4 5 So [sin2x+2(cosx) 2] (1+tanx)=(4 5+2 5) (1+2)=2 5

1)(sinx-4cosx) (5sinx+2cosx)=(2cosx-4cosx) dust omen (10cosx+2cosx)=-2 12=-1 6(2)(sinx) 2+2sinxcosx=4(cosx) 2+4(cosx) 2=8(cosx) 2=8 [1+(tanx) 2]sinx=2cosx tanx=2 original = 8 pie rent[1+2 2]=8 5

1.Using (sinx) 2+(cosx) 2=1, we can see that 2(cosx) 2+3cosxsinx-3(sinx) 2=(cosx) 2+(sinx) 2, so (cosx) 2+3cosxsinx-4(sinx) 2=0Divide the two sides of the equation by (cosx) 2 at the same time, and note that sinx boming pants cosx=tanx gives 1+3tanx-4(tanx) 2=0, so the base Jane has (4t...

The original numerator and denominator are divided by cos 2x at the same time.

1+tan 2x-2tanx) (1-tan 2x)=(1-tanx) 2 limb disturbance((1-tanx)(1+tanx))=1=tanx) (1+tanx)

You copied the question incorrectly. That's the way it works.

ps:sin 2x+cos 2x=1 equation left and right and divide cos 2x to get the front hunger 1 Huishan cos 2x=tan 2x+1So there's the first step.

It can be obtained from [sinx-2cosx][3+2sinx+2cosx]=0.

sinx-2cosx=0

or rent a brother who rolls forward.

sinx+cosx=-3/2

However, since the minimum value of (sinx+cosx) is -root number 2>-3 2, sinx+cosx=-3 2 is rounded.

i.e. sinx-2cosx=0

So sinx=2cosx

So sinx cosx=tanx=2

So 1=(sinx) 2+(cosx) 2=(2cosx) 2+(cosx) 2=5(cosx) 2, so (cosx) 2=1 5

So sin2x=2(sinx)*(cosx)=2(2cosx)*(cosx)=4(cosx) 2=4 5

So [sin2x+2(cosx) 2] (1+tanx)=(4 5+2 5) (1+2)=2

If it is known that (sinx-2cosx)(3+sinx+2cosx)=0, then sinx-2cosx=0 sinx=2cosx or 3+sinx+2cosx=0

Because -1 sinx 1 -1 cosx 1, 3+sinx+2cosx 0

This is true if and only if sinx=-1 cosx=-1, but sinx and cosx cannot =-1 at the same time

Then only sinx=2cosx is true.

Square sin x = 4cos x

1-cos²x=4cos²x

5cos²x=1

cos²x=1/5

sin2x+2cosx) (1+tanx)=(2sinxcosx+2cos x) (1+sinx cosx)=2cosx(sinx+cosx) [(sinx+cosx) cosx].

2cos²x

Knowing that (s-2c)(3+s+2c)=0, then (sin2x+2cos x) (1+tanx) is like this? So.

sin2x+2cos²x)/(1+tanx)=2s * c + 2c²

1+s/c2s * c² +2c³

c+s=2(s+c)c²

s+c=2c²

Then just find the value of cos x.

0=(sinx-2cosx)(3+sinx+2cosx)=>

sinx-2cosx=0 or 3+sinx+2cosx=0 and obviously sx and cx can't get -1 at the same time, so.

sinx-2cosx=0

s-2c=0

s=2cs²=4c²

1-c²=4c²

c²=1/5

So to be asked. 2c²

It can be obtained from [sinx-2cosx][3+2sinx+2cosx]=0.

sinx-2cosx=0 or.

bai sinx+cosx=-3/2

However, because the minimum value du of (sinx+cosx) is -root zhi2>-3 2, daosinx+cosx=-3 2 is rounded.

i.e. sinx-2cosx=0 so version.

weight sinx=2cosx

So sinx cosx=tanx=2

So 1=(sinx) 2+(cosx) 2=(2cosx) 2+(cosx) 2=5(cosx) 2, so (cosx) 2=1 5

So sin2x=2(sinx)*(cosx)=2(2cosx)*(cosx)=4(cosx) 2=4 5

So [sin2x+2(cosx) 2] (1+tanx)=(4 5+2 5) (1+2)=2 5