[munky90]

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

Thanks in advance

-Nathe

[EierVonSatan]

9 sin 2B - 12 sin^3 2B = 3(sin 2B -4sin^3 2B) = 3 sin 6B

[munky90]

I tried that. You end up with



The mark scheme said something about attempting to find a non zero value for sin2B?

[EierVonSatan]

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

[munky90]

Ah, ok, brilliant. Thanks a lot.