Question

what values for, Θ(0≤Θ≤2π) satisfy the equation tan^2Θ=-3/2 sec Θ

Answer 1

its not A

Answer 2

like a first step i think will be helpfully than you rewrite tan theta = sin theta / cos theta

so sqauared what will be ?

Answer 3

i really dont understand how to do this at all....

Answer 4

do you know sec theta what mean exactly ?

Answer 5

Can you answer my question? I take a humor class because I wanna be a television writer. So they gave us this video to see but what is the joke from 12:34 - 12:50? http://www.dailymotion.com/video/x3hrv5o

Answer 6

have you learned about sec cosec ?

Answer 7

no not really this was a random question plugged in our lesson and i just want to know how to do it

Answer 8

but really than you dont learned it again about these nothing so you dont can solving it never you need to know these above wrote what mean and so much again

Answer 9

tan^2 theta = sin^2 theta / cos^2 theta

do you know this ?

Answer 10

yes

Answer 11

looke this please

http://www.math.com/tables/trig/identities.htm

Answer 12

moment please

Answer 13

@highschoolmom2010 please are you here ?

Answer 14

i believ that from above wrote website link you got it easy sec theta equal what ? yes ?

Answer 15

\[\sec ^2 \theta=1+\tan ^2 \theta.\]
make a quadratic in sec theta and solve it

Answer 16

sin^2 theta -3
---------- = ---------- hope you understand how result this
cos^2 theta 2 cos theta

sin^2 theta = 1-cos^2 theta from formula sin^2 x +cos^2 x = 1

and you can divide both sides by 1/cos theta

1-cos^2 theta -3
------------- = ----- cross multiplie
cos theta 2

2(1-cos^2 theta) = -3costheta

2 -2cos^2 theta = -3cos theta sign cos theta = x

2 -2x^2 = -3x

2x^2 -3x -2 = 0 solve this quadratic for x

D = 9 +16 =25

x_1,2 = (-2+/- 5)/4 = -7/4 and 3/4

@Hartmn your opinion about above wrote please ?

Answer 17

@hartnn please

Answer 18

there is nothing wrong in it.

Answer 19

ty

Answer 20

\[\sec ^2 \theta-1=-\frac{ 3 }{ 2 }\sec \theta\]
\[2\sec ^2 \theta+3\sec \theta-2=0\]
\[\frac{ 2 }{ \cos ^2 \theta }+\frac{ 3 }{ \cos \theta }-2=0\]
or \[2+3\cos \theta-2\cos ^2\theta=0\]
or\[2\cos ^2 \theta-3\cos \theta-2=0\]
this is same as yours Mr.Johny

Answer 21

\[2\cos ^2 \theta-4\cos \theta+\cos \theta-2=0\]
\[2\cos \theta \left( \cos \theta-2 \right)+1\left( \cos \theta-2 \right)=0\]
\[\left( \cos \theta-2 \right)\left( 2\cos \theta+1 \right)=0\]
\[\cos \theta=2(rejected)~as~\left| \cos \theta \right|\le1\]
\[\cos \theta=-\frac{ 1 }{ 2 }\]
find theta

Answer 22

\[\cos \theta=-\cos \frac{ \pi }{ 3 }=\cos \left( \pi-\frac{ \pi }{ 3 } \right),\cos \left( \pi+\frac{ \pi }{ 3 } \right)\]
can you find theta now?

Answer 23

hint:
using the definition of the function \(\tan \theta\) and \(\sec \theta\), we can rewrite the equation as follows:
\[\begin{gathered}
\frac{{{{\left( {\sin \theta } \right)}^2}}}{{{{\left( {\cos \theta } \right)}^2}}} + \frac{3}{2}\frac{1}{{\cos \theta }} = 0 \hfill \\
\hfill \\
\frac{{2{{\left( {\sin \theta } \right)}^2} + 3\cos \theta }}{{2{{\left( {\cos \theta } \right)}^2}}} = 0 \hfill \\
\end{gathered} \]
which is equivalent to these conditions:
\[2{\left( {\sin \theta } \right)^2} + 3\cos \theta = 0,\quad \cos \theta \ne 0\]
or these ones:
\[2{\left( {\cos \theta } \right)^2} - 3\cos \theta - 2 = 0,\quad \cos \theta \ne 0\]