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Find tan x

  1. May 26, 2007 #1
    1. The problem statement, all variables and given/known data
    Find tan x if

    [tex]\dfrac {\sin^2 x}{3} + \dfrac {\cos^2 x}{7} = \dfrac {-\sin(2x) + 1}{10}[/tex]


    2. Relevant equations

    Trigonometric identities.

    3. The attempt at a solution

    I have tried removing the cos squared on the LHS by using

    [tex]\cos^2 x + \sin^2 x = 1 [/tex]

    and then using

    [tex] \sin^2 x = \dfrac {1 - \cos(2x)}{2} [/tex]

    then using

    [tex] \cos(2x) = (\cos x + \sin x)(\cos x - \sin x)[/tex]

    Noticing the RHS equals

    [tex](\sin x - \cos x)^2[/tex]

    and factoring with what is now the LHS, would eventually give me an expression similar to

    [tex] a\sin(2x) + b\cos(2x) = c [/tex]

    where a,b,c are constants. If there is now a way to solve for any of sin(2x) or cos(2x), I can then get tan(2x) and hence, tanx.

    This solution has been so long, that it all sounds very suspicious to me. I have a hunch there should be a much easier solution. Thanks for any help, but please, don't post a solution.
     
    Last edited: May 26, 2007
  2. jcsd
  3. May 26, 2007 #2

    VietDao29

    User Avatar
    Homework Helper

    Your method looks okay. But if you are searching for a better one, you may consider dividing both sides by cos2(x), since it's trivial to see that cos(x) = 0 is not a solution to the equation.

    It becomes:
    [tex]\frac{\tan ^ 2 x}{3} + \frac{1}{7} = -2 \frac{\tan x}{10} + \frac{1}{10 \cos ^ 2 (x)}[/tex]

    [tex]\Leftrightarrow \frac{\tan ^ 2 x}{3} + \frac{1}{7} = - \frac{\tan x}{5} + \frac{1}{10} \left( 1 + \tan ^ 2 x \right)[/tex]

    Can you go from here? :) Hint: It's a quadratic.
     
  4. May 26, 2007 #3
    Ouch! That hurts. I didn't even think of dividing by (cosx)^2, otherwise I would have seen it right away. It was right in my face all the time:eek:. If it were a shark, I would be ...
     
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