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

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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
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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..?
ummm 4 sin=-1?
or 2sin=-1?

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

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