anonymous
  • anonymous
what values for θ(0<θ<2pi) satisfy this equation? tan^2 θ = -3/2 secθ
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
this is a poser, hold on i think i have an idea
anonymous
  • anonymous
ok got it ready?
anonymous
  • anonymous
yup!

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anonymous
  • anonymous
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
anonymous
  • anonymous
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
anonymous
  • anonymous
you good from there or you need more steps?
anonymous
  • anonymous
I'm gonna be honest with ya, I have absolutely no idea what I'm doing. So extra steps would be great!
anonymous
  • anonymous
ok but before i write them, do you have any questions about the steps i wrote? there was no real trig, just algebra
anonymous
  • anonymous
the only trig i used was that \(\tan^2(x)=\frac{\sin^2(x)}{\cos^2(x0}\) and \(\sec(x)=\frac{1}{\cos(x)}\)
anonymous
  • anonymous
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
anonymous
  • anonymous
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 \[x=\frac{4\pi}{3}\]
anonymous
  • anonymous
oh my gosh you are wonderful. thank you so much!

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