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# Quick question watch

1. x/2=x-x2 Im trying to find x

Cant work it out, forgotten most of my basic algebra

Any explanation would be most appreciated!
2. x/2=x-x^2

x = 2(x-x^2)

x = 2x - 2x^2

0 = x - 2x^2

x - 2x^2 = 0

x(1 - 2x) = 0

Solve for x.
3. (Original post by SherlockHolmes)
x/2=x-x^2

x = 2(x-x^2)

x = 2x - 2x^2

0 = x - 2x^2

x - 2x^2 = 0

x(1 - 2x) = 0

Solve for x.
so x=0 or x=1/2 ?
4. (Original post by breakeven)
so x=0 or x=1/2 ?
Yes

(Original post by Buttercream:))
anyone been to an interview at university of bradford for pharmacy, wondering what questions they ask really nervous interviews soon!
5. Thanks everyone!
6. (Original post by SherlockHolmes)
Yes

oh awkward -_- lmao! new to this my bad
7. any insight into this question?

sinθtanθ=cosθ+1
Show this can be expressed in the form: 2cos2θ+cosθ-1=0

I took everything over to one side to get sinθtanθ -cosθ-1=0
Then i thought I should use the trig identity to get
sinθ(Sinθ/cosθ)-cosθ-1=0

Kind of stuck from here, do the sinθ's cancel out or is it sin2θ?
Also, what is sinθxcosθ ?
8. (Original post by breakeven)
any insight into this question?

sinθtanθ=cosθ+1
Show this can be expressed in the form: 2cos2θ+cosθ-1=0

I took everything over to one side to get sinθtanθ -cosθ-1=0
Then i thought I should use the trig identity to get
sinθ(Sinθ/cosθ)-cosθ-1=0

Kind of stuck from here, do the sinθ's cancel out or is it sin2θ?
Also, what is sinθxcosθ ?
Multiply through by cosθ then change the sin2θ
9. (Original post by TenOfThem)
Multiply through by cosθ then change the sin2θ
How do I go about doing that?

Im at sinθx(sinθ/cosθ)-cosθ-1=0
Im trying to multiply out the brackets at the minute.
I would get sin2θ+(sinθxcosθ)-cosθ-1=0

Not sure what (sinθxcosθ) is at the minute but I would then change sin2θ into (1-cosθ)
10. (Original post by breakeven)
How do I go about doing that?

Im at sinθx(sinθ/cosθ)-cosθ-1=0
Im trying to multiply out the brackets at the minute.
I would get sin2θ+(sinθxcosθ)-cosθ-1=0

Not sure what (sinθxcosθ) is at the minute but I would then change sin2θ into (1-cosθ)
From here:
sinθx(sinθ/cosθ)-cosθ-1=0

When you multiply out the brackets, you should get sin2θ/cosθ.
11. (Original post by SherlockHolmes)
From here:
sinθx(sinθ/cosθ)-cosθ-1=0

When you multiply out the brackets, you should get sin2θ/cosθ.
So Im at sin2θ/cosθ-cosθ-1=0
Now sure what to do from here, I need to make it the same as 2cos2θ+cosθ-1=0
12. (Original post by breakeven)
So Im at sin2θ/cosθ-cosθ-1=0
Now sure what to do from here, I need to make it the same as 2cos2θ+cosθ-1=0
See TenOfThem's post above.
13. (Original post by SherlockHolmes)
See TenOfThem's post above.
I've tried that but I'm not getting the right answer? Sorry about this, must be driving you crazy but I do appreciate it!
14. (Original post by breakeven)
I've tried that but I'm not getting the right answer? Sorry about this, must be driving you crazy but I do appreciate it!
Post your workings so I can see where you are going wrong
15. (Original post by SherlockHolmes)
Post your workings so I can see where you are going wrong
Man I wish I could just take a picture of my notes haha.

Okay so, sin2θ(sinθ/cosθ)-cosθ-1=0

sin2θ x cosθ -cosθ-1=0
16. (Original post by breakeven)
any insight into this question?

sinθtanθ=cosθ+1
Show this can be expressed in the form: 2cos2θ+cosθ-1=0

I took everything over to one side to get sinθtanθ -cosθ-1=0
Then i thought I should use the trig identity to get
sinθ(Sinθ/cosθ)-cosθ-1=0

Kind of stuck from here, do the sinθ's cancel out or is it sin2θ?
Also, what is sinθxcosθ ?

As I said Multiply by
17. (Original post by TenOfThem)

As I said Multiply by
Oh, god I get it now. Im so sorry!
I wasn't multiplying the 1 by cosθ and i got confused, sorry and thanks so much!
18. (Original post by breakeven)
Man I wish I could just take a picture of my notes haha.

Okay so, sin2θ(sinθ/cosθ)-cosθ-1=0

sin2θ x cosθ -cosθ-1=0
sin^2θ/cosθ - when you multiply by cosθ you just get sin^2]θ.
cosθ + 1 -when you multiply that by cosθ you get cos^2θ + cosθ.
Now using the identity sin^2θ + cos^2θ = 1 you can change the sin^2θ into 1-cos^2θ (take this to the other side)
0=cos^2-1+cos^2θ+cosθ (the original information from this side)
Group like terms 2cos^2θ+cosθ-1

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