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# C2 Trig question watch

1. I just dont get this question:

Solve for x, in the interval 0<x<2π:
cos²x - 6sinx = 5

I got that sinx = 1-cosx? So cos²x - 6(1-cosx) =5???
cos²x-6-cosx=5
cos²x-6cosx+6-5=0
cos²x-6cosx+1=0
and then you use the quadratic formula and solve, as usual? I don't think I did the first bit right, though.
2. (Original post by *girlie*)
I just dont get this question:

Solve for x, in the interval 0<x<2π:
cos²x - 6sinx = 5

I got that sinx = 1-cosx? So cos²x - 6(1-cosx) =5???
cos²x-6-cosx=5
cos²x-6cosx+6-5=0
cos²x-6cosx+1=0
and then you use the quadratic formula and solve, as usual? I don't think I did the first bit right, though.
noooo.
I told you yesterday.
sin²x = 1 - cos²x
NOT sinx = 1 - cosx
3. (Original post by Widowmaker)
noooo.
I told you yesterday.
sin²x = 1 - cos²x
NOT sinx = 1 - cosx
Oops! Yes, I did actually know that. For some reason, i thought it was sin²x
4. So do i convert the cos²x into sin instead? I just can't do trig!
5. (Original post by *girlie*)
So do i convert the cos²x into sin instead? I just can't do trig!
yes, that's right.
6. (Original post by *girlie*)
I just dont get this question:

Solve for x, in the interval 0<x<2π:
cos²x - 6sinx = 5

I got that sinx = 1-cosx? So cos²x - 6(1-cosx) =5???
cos²x-6-cosx=5
cos²x-6cosx+6-5=0
cos²x-6cosx+1=0
and then you use the quadratic formula and solve, as usual? I don't think I did the first bit right, though.
x = 229.81 and 310.19 degrees (2.d.p)
7. Hey, this isn't really in relation to the question, but I was just wondering, is there an "official Edexcel C2 Revision thread"? I'm still fairly new here, and I'm just not really sure about when these things get set up. I'm sure there are a few of us sitting that exam on Monday morning, so it could be useful.
8. Also, how are you meant to do 5sin4θ+1=0 for -90<θ<90?
9. (Original post by *girlie*)
Also, how are you meant to do 5sin4θ+1=0 for -90<θ<90?
sin4theta=-1/5

and work it from that...

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