I was just wondering what anyone else gets for the following question:

solve 0<x<pi

sin 6θ = sin 2θ

P.S. I know that one of the answers is pi over 8

**1**

# Help Required

Started by Bigmanstu, May 19 2004 04:15 PM

6 replies to this topic

### #1

Posted 19 May 2004 - 04:15 PM

### #2

Posted 19 May 2004 - 05:49 PM

The other solutions are 3pi/8, 5pi/8, 7pi/8 (found from drawing a sketch and knowing one solution is pi/8).

I don't really know the proper way of doing these though - it's not something covered in the Higher.

I don't really know the proper way of doing these though - it's not something covered in the Higher.

### #3

Posted 19 May 2004 - 08:39 PM

The proper way of doing things is to put both on the same side ie

sin6&- sin 2& = 0

then solve on from there I think although without knowing & I dont see how that could be done.

sin6&- sin 2& = 0

then solve on from there I think although without knowing & I dont see how that could be done.

Half ideas,half quality, half a million pound law suit!

### #4

Posted 19 May 2004 - 09:23 PM

Thanks guys, I'll give it another go 2moro. I tried it earlier and got

Thanks,

Bigmanstu

*an*answer - not the right 1 but its still an answer! I'll report back when i get it.Thanks,

Bigmanstu

### #5

Posted 19 May 2004 - 09:38 PM

Can I get help on the following question please. I tried it b4 and got the answer but just now, more or less a day b4 the exam, I can't get it!

The answer is

**For the formula S = T10^-kl, find k given that l = 10 when S = 100T.**The answer is

**-0.2**
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### #6

Posted 19 May 2004 - 09:45 PM

Substitute:

100T = T10^(-10k)

100 = 10^(-10k)

log_10 100 = -10k

2 = -10k

k = - 1/5

100T = T10^(-10k)

100 = 10^(-10k)

log_10 100 = -10k

2 = -10k

k = - 1/5

### #7

Posted 19 May 2004 - 09:47 PM

Stupid me!

Thanks a lot for the quick response!

Thanks a lot for the quick response!

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