anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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campbell_st
  • campbell_st
for 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..?
anonymous
  • anonymous
ummm 4 sin=-1?
anonymous
  • anonymous
or 2sin=-1?

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More answers

campbell_st
  • campbell_st
its 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
  • anonymous
yes... so 2 would be tan^2 = -3/2 sec sin/cos ^2 = -3/2 1/cos = sin^2/cos=-3/2 ?
campbell_st
  • campbell_st
well 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..
anonymous
  • anonymous
pi/3 and 5pi/3
anonymous
  • anonymous
??
campbell_st
  • campbell_st
for 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
  • campbell_st
ok... so if you are working in radians the angles are \[\pi + \theta~~and~~~2\pi - \theta\]
anonymous
  • anonymous
Now you lost me completely.
campbell_st
  • campbell_st
remember from an exact value triangle \[\sin(\theta) = \frac{1}{2}~~~then~~~~\theta = \frac{\pi}{6}\]
anonymous
  • anonymous
But there must also be a second one... so pi/6 and 3pi/6?
campbell_st
  • campbell_st
here 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
  • campbell_st
oops 4th quadrant is \[2\pi - \frac{\pi}{6}\]
anonymous
  • anonymous
Still confused. It cant be negative....
anonymous
  • anonymous
pi/6?!?!?!?
anonymous
  • anonymous
and 2pi/6 = pi/3???
campbell_st
  • campbell_st
yes it can here is a graph of the sin curve |dw:1433490149420:dw|
anonymous
  • anonymous
none of my choices are negative though,
campbell_st
  • campbell_st
that is correct.... but some of you choices may include angles that are larger than pi and smaller than 2pi... is that correct...?
anonymous
  • anonymous
this is still for the first one?
campbell_st
  • campbell_st
angles 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
  • anonymous
pi/3, 5pi/3 2pi/3. 4pi/3 pi/6, 3pi,6 7pi/6, 11pi/6
anonymous
  • anonymous
so like d?
campbell_st
  • campbell_st
yes... we are still working on the 1st equation...
campbell_st
  • campbell_st
so like.... that's my guess
campbell_st
  • campbell_st
put it into your calculator does sin(7pi/6) = -1/2 and does sin(11pi/6) = -1/2 try it
anonymous
  • anonymous
I can use the unit circle too, yes?
anonymous
  • anonymous
if so, it is yes for both
campbell_st
  • campbell_st
well there you go... sin can be negative... and the angles are larger then pi radians...
anonymous
  • anonymous
so thats the answer? no more steps?
campbell_st
  • campbell_st
well as I said... that's my best guess... and you told me the unit circle agreed...
anonymous
  • anonymous
so for 2... tan^2 = -3/2sec (sin/cos )^2= -3/2 1/cos ??
campbell_st
  • campbell_st
the unit circle produces the sin curve... that's important
anonymous
  • anonymous
trig is not my best section...
campbell_st
  • campbell_st
well I'd rewrite 2 \[\frac{\sin^2(\theta)}{\cos^2(\theta)} = \frac{-3}{2\cos(\theta)}\] does that make sense..?
campbell_st
  • campbell_st
then multiply both sides by cos^2\[\sin^2(\theta) = \frac{-3 \cos(\theta)}{2}\] is that ok..?
anonymous
  • anonymous
yes.. and then i square root everything?
campbell_st
  • campbell_st
no, 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
  • anonymous
okay.. so cross cancel the cos?
campbell_st
  • campbell_st
no re-write the equation as \[\cos^2(\theta) - \frac{3}{2} \cos(\theta) - 1 = 0\] so you have a quadratic equation you need to solve...
anonymous
  • anonymous
how do i set it up
campbell_st
  • campbell_st
my advice is to use the general quadratic formula \[\cos(\theta) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
anonymous
  • anonymous
but how since theres more than 1?
campbell_st
  • campbell_st
well 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
  • campbell_st
then only consider the value that is less than 1... a solution that is greater than 1 is undefined
campbell_st
  • campbell_st
again, you need to consider where cos is positive... 1st and 4th quadrants... so you will have 2 solutions...
campbell_st
  • campbell_st
hope it made some sense... and good luck with your assessment task.
anonymous
  • anonymous
3.2+/- sqrt -3/2^2+4 ----- 2 so 4.95 and 1.45?
Jack1
  • Jack1
how do u convert from radians to degrees again...?

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