O.K. I'm having a little difficulty with this one and if I could tax you guys I'd like a general point explained.
First the problem. The approach I've tried so far is this:
1.) 1/(sin 3x) + 1/(tan 3x) = 2
2.) 1/(sin 3x) + cos 3x / sin 3x = 2
3.) 1 + cos 3x / six 3x = 2
4.) 1 + cos 3x = 2sin 3x
5.) cos 3x -2sin 3x = -1
6.) let u = 3x
7.) cos u -2sin u = -1
8.) (5^0.5)cos(x + arctan(-2)) = -1
9.) cos(x + arctan(-2)) = -1/(5^0.5)
10.) x + arctan(-2) = arccos(-1/(5^0.5))
11.) x = arccos(-1/(5^0.5)) - arctan(-2)
this gives me 180 degrees, which when I put into the original equation gives me divides by zeroes. I know I can get another cos solution at 10 with 360 - arccos(-1/(5^0.5)), but this also doesn't work. I also tried expansion of the cos 3x and sin 3x, but that didn't seem to get me anywhere.
The general problem I have is to know the conditions when i can multiply and divide trig equations by the functions sin, cos, tan. Did I go wrong way back at 3 when I multipied both sides by sin 3x?
Thanks for your help.