Question

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

Answer 1

this is a poser, hold on i think i have an idea

Answer 2

ok got it
ready?

Answer 3

yup!

Answer 4

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

Answer 5

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

Answer 6

you good from there or you need more steps?

Answer 7

I'm gonna be honest with ya, I have absolutely no idea what I'm doing. So extra steps would be great!

Answer 8

ok but before i write them, do you have any questions about the steps i wrote? there was no real trig, just algebra

Answer 9

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

Answer 10

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

Answer 11

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
\[x=\frac{4\pi}{3}\]

Answer 12

oh my gosh you are wonderful. thank you so much!