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    • Thread Starter
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    This time i thought i got it right...

    Solve the following

     4tan \theta=tan2 \theta............. (0 \leq \theta <360)
    My working..
     4tan \theta=tan2 \theta, tan2 \theta = \frac{ 2tan \theta}{1- tan^2 \theta},
    (\cdot  ^\cdot \cdot)........... \frac{4tan \theta}{2tan \theta} =\frac{1}{1- tan^2 \theta}

      (\cdot  ^\cdot \cdot) ............... 2=\frac{1}{1- tan^2 \theta}....  (\cdot  ^\cdot   \cdot)... 2-2tan^2 \theta = 1[/tex] finaly \pm \sqrt \frac{1}{2}=tan \theta .

    My answers.  35.3, 144.7, 215.3, 324.7

    The books answer 0, 35.3, 144.7, 180, 215.3, 324.7

    the books working the same as mine but then  2tan \theta(1- tan^2 \theta)=tan then..  tan \theta(2 -2tan^2 \theta -1)=0 . How is this possible?
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    You're dividing by tan theta. This is not a good thing to do when tan theta = 0.
    • PS Helper
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    You should avoid dividing through by trig functions (or indeed any function, unless you know for sure that it is never zero). Instead, factorise.

    If you saw the equation x^2=2x you'd take the 2x over and factorise it to give x(x-2)=0; if you divided by x instead then you'd get x=2, which is just one of the two possible solutions. It's exactly the same situation here.
    • Thread Starter
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    I guess thats another lesson learnt, how did they get this?
     tan \theta(2 -2tan^2 \theta -1)=0
    • Thread Starter
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    actualy i se ehow they did this now, thank you both for your help.
 
 
 
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