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Math Help - NEed help solving trig equations

  1. #1
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    NEed help solving trig equations

    I have two trig equations that i just can seem to get. Question is " Solve the trig equations on the given intervals. State any restrictions and give exact answers when posible or round to two
    decimals." First one is:

    cosx=cscxtanx

    and the interval is -\pi\leq x\leq \pi

    second problem is:

    \frac{2tan2x}{1-tan^22x}=\frac{tanx-tan3x}{1+tanxtan3x}

    and the interval is 0\leq x\leq \frac{3\pi}{2}
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  2. #2
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    Re: NEed help solving trig equations

    Have you tried anything yet?

    On the first problem, can we change that csc(x) and tan(x) to something a bit more managable? Perhaps sines and cosines? From there, it should be a bit clearer. On the second, can we factor anything? I always factor whenever I can, and in doing so, a trig identity that I can work with usually becomes apparent. Finally, there is one off the bat that you can work with here.

    Good luck!
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  3. #3
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    Re: NEed help solving trig equations

    Hello, ThisIsFun!

    Solve the trig equations on the given intervals.
    State any restrictions and give exact answers when possible or round to two decimals.

    \frac{2\tan2x}{1-\tan^2\!2x}\:=\:\frac{\tan x-\tan3x}{1+\tan x\tan3x}

    and the interval is: . 0\leq x\leq \tfrac{3\pi}{2}

    The right side has the form: . \tan(A-B) \:=\:\frac{\tan A - \tan B}{1 + \tan A\tan B}

    Hence: . \frac{\tan x-\tan3x}{1+\tan x\tan3x} \:=\:\tan(x-3x) \:=\:\tan(-2x) \;=\;-\tan2x

    The equation becomes: . \frac{2\tan2x}{1-\tan^2\!2x} \;=\;-\tan2x

    . . 2\tan2x \:=\:-\tan2x + \tan^3\!2x \quad\Rightarrow\quad \tan^3\!2x - 3\tan2x \:=\:0

    . . \tan2x(\tan^2\!2x - 3) \:=\:0

    Your turn . . .
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  4. #4
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    Re: NEed help solving trig equations

    Thank You!!!!
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