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

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

Mathematics
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\[\sec ^2\theta-\tan ^2\theta=1,\tan ^2\theta=\sec ^2\theta-1\] \[\sec ^2 \theta-1=-\frac{ 3 }{ 2 }\sec \theta \] \[2 \sec ^2\theta-2+3 \sec \theta=0\] solve further
How did you get the top line?
So \( tan^2 \theta + 1 = sec^2 \theta \)

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Other answers:

it is an identity \[\sin ^2\theta+\cos ^2\theta=1\] divide by \[\cos ^2\theta\] \[\tan ^2\theta+1=\sec ^2\theta\]
Ok going to go solve I think I get it. BRB
solve by quadratic formula or by making factors.
\[2 \sec ^\theta+4 \sec \theta-\sec \theta-2=0\]
correction \[2 \sec ^2\theta\]
Ok I am stuck I have sec = 1/2 and sec = -2
\[2\sec \theta \left( \sec \theta+2 \right)-1\left( \sec \theta+2 \right)=0\] \[\left( \sec \theta+2 \right)\left( 2 \sec \theta-1 \right)=0\] \[\sec \theta=-2,\cos \theta=-\frac{ 1 }{ 2 }=-\cos \frac{ \pi }{ 3 }=\cos \left( \pi \pm \frac{ \pi }{ 3 } \right)=\cos \left( 2n \pi+\pi \pm \frac{ \pi }{ 3 } \right)\] \[\cos \theta=\cos \left\{ \left( 2 n+1 \right)\pi \pm \frac{ \pi }{ 3 } \right\}\] \[\theta=?\] or \[\sec \theta=\frac{ 1 }{ 2 },\cos \theta=2,rejected\] \[as~\left| \cos \theta \right|\le 1\]

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