A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
What values for theta(0< or equal to theta < or equal to 2pi) satisfy the equation?
3sin theta = sin theta  1
tan^2 theta = 3/2sec theta
Help! I have no idea what im doing
anonymous
 one year ago
What values for theta(0< or equal to theta < or equal to 2pi) satisfy the equation? 3sin theta = sin theta  1 tan^2 theta = 3/2sec theta Help! I have no idea what im doing

This Question is Open

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.2for the 1st equation \[3\sin(\theta) = \sin(\theta)  1\] so start by subtracting \[\sin(\theta)\] from both sides of the equation... what do you get..?

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.2its the 2nd option \[2 \sin(\theta) =1\] if you divide both sides of the equation by 2 you get \[\sin(\theta) = \frac{1}{2}\] does that make sense..?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes... so 2 would be tan^2 = 3/2 sec sin/cos ^2 = 3/2 1/cos = sin^2/cos=3/2 ?

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.2well getting back to the 1st equation... you can now solve this... find sin = 1/2 and sin is negative in the 3rd and 4th quadrants so the answer is 180 + theta and 360  theta hope that is ok..

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.2for the 2nd equation rewrite it as \[\frac{\sin^2(\theta)}{\cos^2(\theta)} = \frac{3}{2\cos(\theta)}\] multiply both sides by \[\cos^2(\theta)\] and you get \[\sin^2(\theta) = \frac{3\cos(\theta)}{2}\] then using sin^2 + cos^2 = 1 you can make the substitution \[1  \cos^2(\theta) = \frac{3\cos(\theta)}{2} \] which becomes \[\cos^2(\theta)  \frac{3\cos(\theta)}{2}  1 = 0\] so you can solve the quadratic hope it makes sense

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.2ok... so if you are working in radians the angles are \[\pi + \theta~~and~~~2\pi  \theta\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now you lost me completely.

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.2remember from an exact value triangle \[\sin(\theta) = \frac{1}{2}~~~then~~~~\theta = \frac{\pi}{6}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But there must also be a second one... so pi/6 and 3pi/6?

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.2here is the diagram dw:1433489868047:dw since sin is negative it means the angles are in the 3rd and 4th quadrants... so the 3rd quadrant angle is \[\pi + \frac{\pi}{6}\] and the 4th quadrant angle is \[2\pi  \theta\]

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.2oops 4th quadrant is \[2\pi  \frac{\pi}{6}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Still confused. It cant be negative....

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.2yes it can here is a graph of the sin curve dw:1433490149420:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0none of my choices are negative though,

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.2that is correct.... but some of you choices may include angles that are larger than pi and smaller than 2pi... is that correct...?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is still for the first one?

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.2angles between pi and 2 pi... return negative values... for sin e.g. find sin(3pi/2) on your calculator... it will return a value of 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0pi/3, 5pi/3 2pi/3. 4pi/3 pi/6, 3pi,6 7pi/6, 11pi/6

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.2yes... we are still working on the 1st equation...

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.2so like.... that's my guess

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.2put it into your calculator does sin(7pi/6) = 1/2 and does sin(11pi/6) = 1/2 try it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I can use the unit circle too, yes?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if so, it is yes for both

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.2well there you go... sin can be negative... and the angles are larger then pi radians...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so thats the answer? no more steps?

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.2well as I said... that's my best guess... and you told me the unit circle agreed...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so for 2... tan^2 = 3/2sec (sin/cos )^2= 3/2 1/cos ??

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.2the unit circle produces the sin curve... that's important

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0trig is not my best section...

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.2well I'd rewrite 2 \[\frac{\sin^2(\theta)}{\cos^2(\theta)} = \frac{3}{2\cos(\theta)}\] does that make sense..?

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.2then multiply both sides by cos^2\[\sin^2(\theta) = \frac{3 \cos(\theta)}{2}\] is that ok..?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes.. and then i square root everything?

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.2no, make a substitution sin^2 = 1  cos^2 this come from sin^2 + cos^2 = 1 so you get \[1  \cos^2(\theta) = \frac{3\cos(\theta)}{2}\] is that ok

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay.. so cross cancel the cos?

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.2no rewrite the equation as \[\cos^2(\theta)  \frac{3}{2} \cos(\theta)  1 = 0\] so you have a quadratic equation you need to solve...

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.2my advice is to use the general quadratic formula \[\cos(\theta) = \frac{b \pm \sqrt{b^2  4ac}}{2a}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but how since theres more than 1?

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.2well the coefficients of the quadratic are a = 1, b = 3/2 and c = 1 if you struggled with trig you must be good at solving quadratics

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.2then only consider the value that is less than 1... a solution that is greater than 1 is undefined

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.2again, you need to consider where cos is positive... 1st and 4th quadrants... so you will have 2 solutions...

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.2hope it made some sense... and good luck with your assessment task.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.03.2+/ sqrt 3/2^2+4  2 so 4.95 and 1.45?

Jack1
 one year ago
Best ResponseYou've already chosen the best response.0how do u convert from radians to degrees again...?
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.