Question

What values for θ (0≤θ≤2π) satisfy the equation?

Answer 1

\[tan^2\theta=-\frac{3}{2}sec\theta\]

Answer 2

sorry i suck at trig :(

Answer 3

It's all good :)

Answer 4

do U know the identity that relates both of the \[\sin (\Theta) , \tan (\Theta)\] ?

Answer 5

\[\sec ^2 \theta -\tan ^2 \theta=1\]

Answer 6

this is how to start

\[\tan^2(\theta) = \frac{\sin^2(\theta)}{\cos^2(\theta)}\]

and

\[\sec(\theta) = \frac{1}{\cos(\theta)}\]

multiply both sides of the equation by cos^2

\[\sin^2(\theta) = \frac{3}{2} \cos(\theta)\]

now use the trig identity sin^2 + cos^2 = 1

so

\[1 - \cos^2(\theta) = \frac{3}{2} \cos(\theta)\]

which can be written as

\[\cos^2(\theta) + \frac{3}{2}\cos(\theta) - 1 = 0\]

so now you have a quadratic that can be solved...

hope it helps

Answer 7

\[\tan ^2 \theta=\sec ^2 \theta-1\]

Answer 8

\[\sec ^2 \theta -1=\frac{ 3 }{ 2 }\cos \theta\]
\[2 \sec ^2 \theta-2=3 \sec \theta\]
solve for sec theta

Answer 9

when you solve the quadratic using the general quadratic formula you are solving

\[\cos(\theta) = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}\]

this will give an exact value, then find the angle measures

Answer 10

$\theta$

Answer 11

$tan^2$

Answer 12

oops should be will give 2 exact values... this will allow you to find the angles...

hope it helps

Answer 13

So , I'll assume you don't know it :D

\[\tan^2 (\Theta) +1 = \sec^2 (\Theta)\]
so nw U have \[\tan^2 (\Theta) = \sec^2 (\Theta) -1\]

Simply you plug this into you equation and you will get sth like this :
\[\sec^2 (\Theta) -1 = -\frac{ 3 }{ 2 } \sec(\Theta)\]

You can easily get the \[-\frac{ 3 }{ 2 } \sec(\Theta)\] to the right hand side by changing the sign :)

The you will get :
\[\sec^2(\Theta) + 1.5 \sec(\Theta) - 1 = 0\]

nw guess what !

It's a quadratic equation that could be solved easily Ur calculator or the formula and the you will get
\[\sec(\Theta) = \frac{ 1 }{ 2 } , -2\]

This is for sec to get theta you have to flip the answer and you will get \[\cos(\Theta) = -\frac{ 1 }{ 2 } , 2\]
So the second one will be rejected as 2 is not a value for any Theta !

nw you will get Theta simply like this : \[\Theta = \cos^{-1} (-\frac{ 1 }{ 2 }) = 120^o = \frac{ 2 }{ 3 }π\] radians :D


Let me know if you got it :)

Answer 14

Ah! sorry...was afk :(

Answer 15

\[\sec ^2 \theta-1=-\frac{ 3 }{ 2 }\sec \theta\]
\[2\sec ^2\theta+3\sec \theta-2=0\]
\[2\sec ^2\theta+4\sec \theta-\sec \theta-2=0\]
\[2\sec \theta \left( \sec \theta+2 \right)-1\left( \sec \theta+2 \right)=0\]
\[\left( \sec \theta+2 \right)\left( 2\sec \theta-1 \right)=0\]
\[Either~\sec \theta+2=0,\sec \theta=-2,\cos \theta=-\frac{ 1 }{ 2 }=-\cos \frac{ \pi }{ 3 }=\cos \left( \pi \pm \frac{ \pi }{ 3 } \right)\]
\[\theta=\pi \pm \frac{ \pi }{ 3 }\]
\[or~2\sec \theta-1=0.\sec \theta=\frac{ 1 }{ 2 },\cos \theta=2\]
rejected as\[\left| \cos \theta \right|\le 1\]