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

1. anonymous

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

2. anonymous

3. anonymous

yup!

4. 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

5. 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

6. anonymous

you good from there or you need more steps?

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. 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

11. 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 $x=\frac{4\pi}{3}$

12. anonymous

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