Results 1 to 4 of 4

Math Help - 1/(sin 3x) + 1/(tan 3x) = 2

  1. #1
    Newbie
    Joined
    Oct 2011
    Posts
    5

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,740
    Thanks
    645

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

    Hello, HumanShale!

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


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

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

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

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


    \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]:

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

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

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

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

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

    . . . . . . . . . . . . . . . . . . x \:=\:\tfrac{1}{3}\left[\arcsin\left(\tfrac{1}{\sqrt{5}}\right) + \arctan\left(\tfrac{1}{2}\right)\right]
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,569
    Thanks
    1428

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

    An easier method perhaps?

    \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 \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...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2011
    Posts
    5

    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.
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum