TRIGONOMETRY HELP

Question

solve, for 0<x<90°,

4 cos 3x − 1 = 2 sin 3x tan 3x

preferably using quadratic

think

please help would be appreciated

Answer 1

Fourth line should be

4 cos² 3x - cos 3x = 2 sin² 3x

4 cos² 3x - cos 3x = 2(1 - cos² 3x\

4 cos² 3x - cos 3x = 2 - 2 cos² 3x

6 cos² 3x - cos 3x - 2 = 0

Use u substitution or the Quadratic Formula

Answer 2

4 cos 3x − 1 = 2 sin 3x tan 3x

4cos3x-1 = 2sin3x(sin3x/cos3x)

" = 2(sin3x)^2/cos3x

4(cos3x)^2-1=2(sin3x)^2

2(cos3x)-1 =sin3x

The best I can do is x=12.33°=-.0041945.......