How Do You Solve Trigonometric Equations Involving Sin, Cos, and Tan?

[Nope]

Homework Statement

Solve 2sinx+1=tanx 0\leqx\prec2 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, please help!
Thanks

Btw , can anyone show me the step for this one
2cotxsinx+cotx-2sinx-1=0 domain:0\leqx\prec2 pi
thanks!

 

[rock.freak667]

What is sin2x-sinx equal to?

(Hint: Sum to product formula)

 

[Nope]

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

 

[rock.freak667]

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

 

[Nope]

yes that will work as well

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??

 

[Mark44]

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

 

[Nope]

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..

 

[Mark44]

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.