## haleyelizabeth2017 one year ago What values for θ (0≤θ≤2π) satisfy the equation?

1. haleyelizabeth2017

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

2. acxbox22

sorry i suck at trig :(

3. haleyelizabeth2017

It's all good :)

4. anonymous

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

5. anonymous

$\sec ^2 \theta -\tan ^2 \theta=1$

6. campbell_st

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

7. anonymous

$\tan ^2 \theta=\sec ^2 \theta-1$

8. anonymous

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

9. campbell_st

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

10. anonymous

$\theta$

11. anonymous

$tan^2$

12. campbell_st

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

13. anonymous

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 :)

14. haleyelizabeth2017

Ah! sorry...was afk :(

15. anonymous

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