C3 Trigonometry question

I'm having so much trouble with these big questions :frown:

This one is 6 marks

how do you solve

3sin6xcosec2x = 4

NB: the original question has beta instead of x

i can't thank you guys enough. :redface:

Reply 1

rayman198

You need to use the patterns from part i) and ii)

sin 3x = 3sinx - 4sin^3 x (using x's instead of thetas)

therefore 3sin 6Bcosec2B = 3(sin(2B))cosec2B = 3 (3sin 2B - 4sin^3 2B)(1/sin^2B) (im using B's here instead of betas)(cosec 2B changed to 1/sin^2B in final bracket)

expand to get:

9 - 12sin^2 2B = 4

5/12 = sin^2 2B
sin 2B = sqrt 5/12

(adjust domain to 0<2B<180) and solve, then halve the answers to provide solutions for B not 2B.

Reply 2

icecreamman

doesnt cosec2B become 1/sin2B rather than 1/sin^2B?

Reply 3

john-boro

thats what he means I think. It works if you use 1/sin2B

Reply 4

showdown_counts

3sin6x.cosex2x=4
3six6x(1/sin2x)=4
3sin6x=4sin2x
3sin(4x+2x)=4sin(4x-2x)
3(sin4x.cos2x+cos4x.sin2x)=4(sin4x.cos2x-cos4x.sin2x)

**do all the equations**

sin4x.cos2x=7cos4x.sin2x

(sin4x/cos4x)=7(sin2x/cos2x)
tan4x=7tan2x
tan(2x+2x)=7tan2x
(tan2x+tan2x)/(1-tan^2.2x)=7tan2x
2tan2x/tan2x=7(1-tan^2.2x)
7-7tan^2.2x=2
tan^2.2x=5/7
2x= 40.2, 139.8, 220.2, 319.8, 400.2, 499.2, 580.2, 679.2
x=20.1, 69.9, 110.1, 159.9, 200.1, 249.6, 290.1, 339.6 ( arrange it according to limits)

I dont know if its correct or not...but i tried my best.