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infinitemoes
Group Title
what values for θ(0<θ<2pi) satisfy this equation? tan^2 θ = 3/2 secθ
 one year ago
 one year ago
infinitemoes Group Title
what values for θ(0<θ<2pi) satisfy this equation? tan^2 θ = 3/2 secθ
 one year ago
 one year ago

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satellite73 Group TitleBest ResponseYou've already chosen the best response.1
this is a poser, hold on i think i have an idea
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
ok got it ready?
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
you good from there or you need more steps?
 one year ago

infinitemoes Group TitleBest ResponseYou've already chosen the best response.0
I'm gonna be honest with ya, I have absolutely no idea what I'm doing. So extra steps would be great!
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
ok but before i write them, do you have any questions about the steps i wrote? there was no real trig, just algebra
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
the only trig i used was that \(\tan^2(x)=\frac{\sin^2(x)}{\cos^2(x0}\) and \(\sec(x)=\frac{1}{\cos(x)}\)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
now we have \[2(1\cos^2(x))+3\cos(x)=0\] which is like solving \[2(1u^2)+3u=0\] rewrite as \[22u^3+3u=0\] or \[2u^23u2=0\] factor as \[(2u+1)(u2)=0\] 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
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
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 \[x=\frac{4\pi}{3}\]
 one year ago

infinitemoes Group TitleBest ResponseYou've already chosen the best response.0
oh my gosh you are wonderful. thank you so much!
 one year ago
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