Question

Solve Trig Equation


\( tan^2 \theta = -\frac{3}{2} sec \theta \)

Answer 1

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

Answer 2

How did you get the top line?

Answer 3

So \( tan^2 \theta + 1 = sec^2 \theta \)

Answer 4

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

Answer 5

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

Answer 6

solve by quadratic formula or by making factors.

Answer 7

\[2 \sec ^\theta+4 \sec \theta-\sec \theta-2=0\]

Answer 8

correction
\[2 \sec ^2\theta\]

Answer 9

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

Answer 10

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