1/(sin 3x) + 1/(tan 3x) = 2

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.

Re: 1/(sin 3x) + 1/(tan 3x) = 2

Hello, HumanShale!

I don't understand step (8).

I'll show you my approach.

We have: .

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. [1]

Substitute into [1]:

. .

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. . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . .

Re: 1/(sin 3x) + 1/(tan 3x) = 2

An easier method perhaps?

And now it's a case of having to check for extraneous roots because you had to square the equation originally.

Hmm, maybe this wasn't easier after all...

Re: 1/(sin 3x) + 1/(tan 3x) = 2

Thanks to both Soroban and Prove It for your replies. Just to explain step 8 in my post:

the idea was that Acos x - Bsin x = Rcos(x + t) where R = (A^2 + B^2)^0.5 and t = arctan(B/A).

Thanks to your posts I was able to go back through it and see the errors I was making - arctan (-2) should be arctan (2). I also wrote it up really badly and forgot to keep using the u I replaced 3x with.

It can be solved using the cos identity to get a solution at: x = (arccos(-1/(5^0.5)) - arctan(2)) / 3. Then add / subtract 360 degrees. That was what I was trying to do.

I liked Prove It's approach as well - I messed around with some 3x expansions but didn't think to square both sides.