ocr maths c3 help?

Cosec = 1/sin

Bring the 4 over and you get 3sin.1/sin = 4

One of the sin will cancel out the 1/sin so your left with 2sin6b x 2b = 4

That's as far as I can take it without fully working it out which I CBA to do, try that

how do you solve this equation ?

3sin6bcosec2b = 4 for 0 <b <90degrees

please state every step so that i know how to do this question

I only reached up to 7cos(4b)sin(2b)-sin(4b)cos(2b)

Cosec = 1/sin

Bring the 4 over and you get 3sin.1/sin = 4

One of the sin will cancel out the 1/sin so your left with 2sin6b x 2b = 4

That's as far as I can take it without fully working it out which I CBA to do, try that

It needs a lot of good manipulation (or someone could show if it's actually simpler than I think it is).
$3\sin{(6b)}\text{cosec}{(2b)} = 4 \implies 3\sin(6b) = 4\sin(2b)$

Now, nothing useful can be seen, until you actually find the common factors. Note that:
$\sin(6b) = \sin(4b+2b) = \sin(4b)cos(2b)+\cos(4b)\sin(2b)$ and $\sin(2b) = \sin(4b-2b) = \sin(4b)cos(2b)-\cos(4b)\sin(2b)$.

I think you could work from there. If you're stuck still, ask.