## anonymous one year ago Solve Trig Equation $$tan^2 \theta = -\frac{3}{2} sec \theta$$

1. anonymous

$\sec ^2\theta-\tan ^2\theta=1,\tan ^2\theta=\sec ^2\theta-1$ $\sec ^2 \theta-1=-\frac{ 3 }{ 2 }\sec \theta$ $2 \sec ^2\theta-2+3 \sec \theta=0$ solve further

2. anonymous

How did you get the top line?

3. anonymous

So $$tan^2 \theta + 1 = sec^2 \theta$$

4. anonymous

it is an identity $\sin ^2\theta+\cos ^2\theta=1$ divide by $\cos ^2\theta$ $\tan ^2\theta+1=\sec ^2\theta$

5. anonymous

Ok going to go solve I think I get it. BRB

6. anonymous

solve by quadratic formula or by making factors.

7. anonymous

$2 \sec ^\theta+4 \sec \theta-\sec \theta-2=0$

8. anonymous

correction $2 \sec ^2\theta$

9. anonymous

Ok I am stuck I have sec = 1/2 and sec = -2

10. anonymous

$2\sec \theta \left( \sec \theta+2 \right)-1\left( \sec \theta+2 \right)=0$ $\left( \sec \theta+2 \right)\left( 2 \sec \theta-1 \right)=0$ $\sec \theta=-2,\cos \theta=-\frac{ 1 }{ 2 }=-\cos \frac{ \pi }{ 3 }=\cos \left( \pi \pm \frac{ \pi }{ 3 } \right)=\cos \left( 2n \pi+\pi \pm \frac{ \pi }{ 3 } \right)$ $\cos \theta=\cos \left\{ \left( 2 n+1 \right)\pi \pm \frac{ \pi }{ 3 } \right\}$ $\theta=?$ or $\sec \theta=\frac{ 1 }{ 2 },\cos \theta=2,rejected$ $as~\left| \cos \theta \right|\le 1$