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Math Help - Solving trigonometric equations

  1. #1
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    Solving trigonometric equations

    2Sec(x)*tan(x)+sec(x)^2=0

    I know that the answer is 210 degrees from using my graph. How would I do this by hand?

    I tried this.

    sec(x)(2tan(x)+sec(x))=0

    I get stuck here I have also changed the sec(x)^2 to Tan(x)^2+1. But then I get stuck because I still have two different trigonometric functions. Please help. Thanks
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  2. #2
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    sec(2tan+sec)=0
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  3. #3
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    what does that mean? is that the one I want to use? how do i solve the the 2tan+sec if they are different?
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  4. #4
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    Quote Originally Posted by IDontunderstand View Post
    what does that mean? is that the one I want to use? how do i solve the the 2tan+sec if they are different?
    How would you solve x(2y+x)=0?
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  5. #5
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    \displaystyle 2\sec(x)\tan(x)+\sec^2(x)=0

    \displaystyle \frac{2}{\cos(x)}\frac{\sin(x)}{\cos(x)}+\frac{1}{  \cos^2(x)}=0

    \displaystyle \frac{2\sin(x)}{\cos^2(x)}+\frac{1}{\cos^2(x)}=0

    \displaystyle \frac{2\sin(x)+1}{\cos^2(x)}=0

    Can you finish it?
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  6. #6
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    I don't know I have two variables and one equation.
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  7. #7
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    Yes, but if the left-hand side is zero, what is the numerator?
    Double-check your denominator will also not be zero for the solution x.
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  8. #8
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    Quote Originally Posted by Archie Meade View Post
    Yes, but if the left-hand side is zero, what is the numerator?
    Double-check your denominator will also not be zero for the solution x.
    zero so x can be zero and y can thus be hafl the opposite of x
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  9. #9
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    yes multiple each side by cos^2 so that I get rid of that. Then solve sinx to be -1/2 and get the angle of 210 and 330.
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  10. #10
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    Yes
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  11. #11
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    Quote Originally Posted by Archie Meade View Post
    In "pickslides" post, the numerator can is zero.
    If we say "x" is secant whih is zero, then I have to find arcsecant of zero, which is undefinded.
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  12. #12
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    Quote Originally Posted by IDontunderstand View Post
    If we say "x" is secant whih is zero, then I have to find arcsecant of zero, which is undefinded.
    For your 2 solutions x,

    cosx\ \ne\ 0

    so those solutions are fine, you don't have 0/0
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  13. #13
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    Great solution. Thanks
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