1. Nov 23, 2009

### Nope

1. The problem statement, all variables and given/known data
Solve 2sinx+1=tanx 0$$\leq$$x$$\prec$$2 pi

2. Relevant equations

3. The attempt at a solution
I tried to put it in this way
2sinx+1=tanx
2sinx+1=(sinx/cosx)
2sinxcosx+cosx-sinx=0
sin2x+cosx-sinx=0
I don't know what to do next, I am stuck, plz help!
Thanks

Btw , can anyone show me the step for this one
2cotxsinx+cotx-2sinx-1=0 domain:0$$\leq$$x$$\prec$$2 pi
thanks!

2. Nov 23, 2009

### rock.freak667

What is sin2x-sinx equal to?

(Hint: Sum to product formula)

3. Nov 24, 2009

### Nope

I haven't learn that yet...
Sin(a+b),cos(a+b)...
Sin2A=2sinAcosA, cos2A=cosx^2-sin^2....
and sinx^2+cosx^2=1 ....and other form
this is all i know
We just finished this unit today, i don't think i will learn that...

Can i square both side?
sin2x+cosx-sinx=0
(sin2x)^2=(sinx-cosx)^2
(sin2x)^2=sinx^2-2cosxsinx+cosx^2
(sin2x)^2=1-sin2x
sin2x^x+sin2x-1=0

Last edited: Nov 24, 2009
4. Nov 24, 2009

### rock.freak667

yes that will work as well

5. Nov 25, 2009

### Nope

http://www.wolframalpha.com/input/?i=2sinx+1-tanx=0
http://www.wolframalpha.com/input/?i=sin2x^2%2Bsin2x-1%3D0
But it seem that both graph is different...
where did i go wrong??

6. Nov 25, 2009

### Staff: Mentor

What did you get when you solved this equation?
sin2x^x+sin2x-1=0

7. Nov 25, 2009

### Nope

i got 4 solution,
but 2sinx+1=tanx only have 2 solution..

8. Nov 25, 2009

### Staff: Mentor

Squaring both sides of an equation often introduces extraneous solutions that aren't solutions of the original, unsquared equation. Check each of your four solutions in your original equation. Keep only the ones that satisfy it.