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# OCR January 2006 C3 q9 part iii Tweet

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1. OCR January 2006 C3 q9 part iii
Hey. I got really unstuck with this question so please could someone give me some help?

9,iii) Solve, for 0<B<90, the equation

I managed to get down to and earlier in the question I had to prove that

it might be clearer if you see the paper here

-Nathe
2. Re: OCR January 2006 C3 q9 part iii
9 sin 2B - 12 sin^3 2B = 3(sin 2B -4sin^3 2B) = 3 sin 6B
3. Re: OCR January 2006 C3 q9 part iii
I tried that. You end up with

The mark scheme said something about attempting to find a non zero value for sin2B?
4. Re: OCR January 2006 C3 q9 part iii
3(sin 2B -4sin^3 2B)cosec 2B = 3 (1 - 4sin^2 2B)

divided by sin 2B, so you have to solve sin 2B = 0 also
5. Re: OCR January 2006 C3 q9 part iii
Ah, ok, brilliant. Thanks a lot.