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

• Mar 30th 2013, 05:20 AM
HumanShale
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?

• Mar 30th 2013, 06:33 AM
Soroban
Re: 1/(sin 3x) + 1/(tan 3x) = 2
Hello, HumanShale!

I don't understand step (8).
I'll show you my approach.

Quote:

$\displaystyle \frac{1}{\sin 3x} + \frac{1}{\tan 3x} \:=\:2$

We have: .$\displaystyle \frac{1}{\sin3x} + \frac{\cos3x}{\sin3x} \:=\:2 \quad\Rightarrow\quad \frac{1+\cos3x}{\sin3x} \:=\:2$

. . . . . . . .$\displaystyle 1 + \cos3x \:=\:2\sin3x \quad\Rightarrow\quad 2\sin3x - \cos3x \:=\:1$

$\displaystyle \text{Divide by }\sqrt{5}\!:\;\;\tfrac{2}{\sqrt{5}}\sin3x - \tfrac{1}{\sqrt{5}}\cos3x \:=\:\tfrac{1}{\sqrt{5}}$ . [1]

$\displaystyle \text{Let }\theta \,=\,\arctan\tfrac{1}{2} \quad\Rightarrow\quad \sin\theta \,=\,\tfrac{1}{\sqrt{5}} \quad\Rightarrow\quad \cos\theta \,=\,\tfrac{2}{\sqrt{5}}$

Substitute into [1]:

. . $\displaystyle \cos\theta\sin3x - \sin\theta\cos3x \:=\:\tfrac{1}{\sqrt{5}}$

. . . . . . . . . . .$\displaystyle \sin(3x-\theta) \:=\:\tfrac{1}{\sqrt{5}}$

. . . . . . . . . . . . . .$\displaystyle 3x-\theta \:=\:\arcsin\left(\tfrac{1}{\sqrt{5}}\right)$

. . . . . . . . . . . . . . . . .$\displaystyle 3x \:=\:\arcsin\left(\tfrac{1}{\sqrt{5}}\right) + \theta$

. . . . . . . . . . . . . . . . . .$\displaystyle x \:=\:\tfrac{1}{3}\left[\arcsin\left(\tfrac{1}{\sqrt{5}}\right) + \theta\right]$

. . . . . . . . . . . . . . . . . .$\displaystyle x \:=\:\tfrac{1}{3}\left[\arcsin\left(\tfrac{1}{\sqrt{5}}\right) + \arctan\left(\tfrac{1}{2}\right)\right]$
• Mar 30th 2013, 07:20 AM
Prove It
Re: 1/(sin 3x) + 1/(tan 3x) = 2
An easier method perhaps?

\displaystyle \displaystyle \begin{align*} \frac{1}{\sin{(3x)}} + \frac{1}{\tan{(3x)}} &= 2 \\ \frac{1}{\sin{(3x)}} + \frac{\cos{(3x)}}{\sin{(3x)}} &= 2 \\ \frac{1 + \cos{(3x)}}{\sin{(3x)}} &= 2 \\ 1 + \cos{(3x)} &= 2\sin{(3x)} \\ \left[ 1 + \cos{(3x)} \right] ^2 &= \left[ 2\sin{(3x)} \right] ^2 \\ 1 + 2\cos{(3x)} + \cos^2{(3x)} &= 4\sin^2{(3x)} \\ 1 + 2\cos{(3x)} + \cos^2{(3x)} &= 4 \left[ 1 - \cos^2{(3x)} \right] \\ 1 + 2\cos{(3x)} + \cos^2{(3x)} &= 4 - 4\cos^2{(3x)} \\ 5\cos^2{(3x)} + 2\cos{(3x)} - 3 &= 0 \\ 5\cos^2{(3x)} + 5\cos{(3x)} - 3\cos{(3x)} - 3 &= 0 \\ 5\cos{(3x)} \left[ \cos{(3x)} + 1 \right] - 3 \left[ \cos{(3x)} + 1 \right] &= 0 \end{align*}

\displaystyle \displaystyle \begin{align*} \left[ \cos{(3x)} + 1 \right] \left[ 5\cos{(3x)} - 3 \right] &= 0 \\ \cos{(3x)} + 1 = 0 \textrm{ or } 5\cos{(3x)} - 3 &= 0 \\ \cos{(3x)} = -1 \textrm{ or } \cos{(3x)} &= \frac{3}{5} \\ 3x = \pi + 2\pi n \textrm{ or } 3x &= \left\{ \arccos{\left( \frac{3}{5} \right) } , 2\pi - \arccos{ \left( \frac{3}{5} \right) } \right\} + 2\pi n \textrm{ where } n \in \mathbf{Z} \\ x &= \left\{ \frac{1}{3} \arccos{ \left( \frac{3}{5} \right)} , \frac{ \pi }{3} , \frac{ 2\pi }{3} - \frac{1}{3} \arccos{ \left( \frac{3}{5} \right) } \right\} + \frac{2\pi n}{3} \end{align*}

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...
• Mar 30th 2013, 07:46 AM
HumanShale
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.