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infinitemoes

  • 2 years ago

what values for θ(0<θ<2pi) satisfy this equation? tan^2 θ = -3/2 secθ

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  1. satellite73
    • 2 years ago
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    this is a poser, hold on i think i have an idea

  2. satellite73
    • 2 years ago
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    ok got it ready?

  3. infinitemoes
    • 2 years ago
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    yup!

  4. satellite73
    • 2 years ago
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    it is not that hard, just my first attempt didn't work add \(\frac{3}{2}\sec(x)\) to start withi \[\tan^2(x)+\frac{3}{2}\sec(x)=0\] then rewrite as \[\frac{\sin^2(x)}{\cos^2(x)}+\frac{3}{2\cos(x)}=0\] add up to get \[\frac{2\sin^2(x)+3\cos(x)}{2\cos^2(x)}=0\] then set the numerator equal to zero and solve

  5. satellite73
    • 2 years ago
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    you get \[2\sin^2(x)+3\cos(x)=0\] rewrite as \[2(1-\cos^2(x))+3\cos(x)=0\] and solve the quadratic equation in cosine

  6. satellite73
    • 2 years ago
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    you good from there or you need more steps?

  7. infinitemoes
    • 2 years ago
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    I'm gonna be honest with ya, I have absolutely no idea what I'm doing. So extra steps would be great!

  8. satellite73
    • 2 years ago
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    ok but before i write them, do you have any questions about the steps i wrote? there was no real trig, just algebra

  9. satellite73
    • 2 years ago
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    the only trig i used was that \(\tan^2(x)=\frac{\sin^2(x)}{\cos^2(x0}\) and \(\sec(x)=\frac{1}{\cos(x)}\)

  10. satellite73
    • 2 years ago
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    now we have \[2(1-\cos^2(x))+3\cos(x)=0\] which is like solving \[2(1-u^2)+3u=0\] rewrite as \[2-2u^3+3u=0\] or \[2u^2-3u-2=0\] factor as \[(2u+1)(u-2)=0\] so \[u=-\frac{1}{2}\] or \[u=2\] i.e. \[\cos(x)=-\frac{1}{2}\] which is the only solution, because cosine cannot be 2

  11. satellite73
    • 2 years ago
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    then look in the unit circle to see for what values of \(x\) you get \[\cos(x)=-\frac{1}{2}\] and you will see it is \[x=\frac{2\pi}{3}\] or \[x=\frac{4\pi}{3}\]

  12. infinitemoes
    • 2 years ago
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    oh my gosh you are wonderful. thank you so much!

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