How do you solve 3sin6βcosec2β = 4, in the range 0<β<90?

Question

Previous details given were
sin 3θ = 3sinθ - 4sin^3θ

Answer 1

3 sin(6β)/sin(2β) =
= 3 [3 sin(2β) - 4 sin^3(2β)]/sin(2β)=
= 9 - 12 sin^2(2β) = 4

Then

sin(2β) = sqrt(5/12)

or

β = (1/2) arcsin(sqrt(5/12)) = 0.3508 rad

Answer 2

sin 3θ = 3sinθ - 4sin^3θ
So: Let 6β = 3Θ
and 2β = Θ
and we get:
sin 6β = 3sin2β - 4sin^3(2β)
Using this to solve:
3sin6βcosec2β = 4
3(3sin2β - 4sin^3(2β))*1/sin2β = 4
9 - 12sin^2(2β)=4
sin^2(2β) = 5/12
sin 2β = ± .6455
the general solution in degrees for 2β is:
2β = 24.62 ± 180n, -24.62 ± 180n
β = 12.31 ± 90n, -12.31 ± 90n
since your range is (0, 90)
β = 12.31, 77.69

Answer 3

is that even a real question ??????

Answer 4

i have no idea!!!

Answer 5

oh...