PLEASE HELP, need this today

  • Thread starter Nope
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  • #1
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Homework Statement


Solve 2sinx+1=tanx 0[tex]\leq[/tex]x[tex]\prec[/tex]2 pi

Homework Equations





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[tex]\leq[/tex]x[tex]\prec[/tex]2 pi
thanks!
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
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What is sin2x-sinx equal to?

(Hint: Sum to product formula)
 
  • #3
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What is sin2x-sinx equal to?

(Hint: Sum to product formula)

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
and then use quadratic
 
Last edited:
  • #4
rock.freak667
Homework Helper
6,223
31
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
and then use quadratic

yes that will work as well
 
  • #6
35,231
7,053
What did you get when you solved this equation?
sin2x^x+sin2x-1=0
 
  • #7
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What did you get when you solved this equation?
sin2x^x+sin2x-1=0

i got 4 solution,
but 2sinx+1=tanx only have 2 solution..
 
  • #8
35,231
7,053
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.
 

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