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

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- 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

ummm 4 sin=-1?

- anonymous

or 2sin=-1?

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## More answers

- 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

yes...
so 2 would be
tan^2 = -3/2 sec
sin/cos ^2 = -3/2 1/cos = sin^2/cos=-3/2 ?

- 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

pi/3 and 5pi/3

- anonymous

??

- 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

ok... so if you are working in radians
the angles are
\[\pi + \theta~~and~~~2\pi - \theta\]

- anonymous

Now you lost me completely.

- campbell_st

remember from an exact value triangle
\[\sin(\theta) = \frac{1}{2}~~~then~~~~\theta = \frac{\pi}{6}\]

- anonymous

But there must also be a second one... so pi/6 and 3pi/6?

- 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

oops 4th quadrant is
\[2\pi - \frac{\pi}{6}\]

- anonymous

Still confused. It cant be negative....

- anonymous

pi/6?!?!?!?

- anonymous

and 2pi/6 = pi/3???

- campbell_st

yes it can here is a graph of the sin curve |dw:1433490149420:dw|

- anonymous

none of my choices are negative though,

- 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

this is still for the first one?

- 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

pi/3, 5pi/3
2pi/3. 4pi/3
pi/6, 3pi,6
7pi/6, 11pi/6

- anonymous

so like d?

- campbell_st

yes... we are still working on the 1st equation...

- campbell_st

so like.... that's my guess

- campbell_st

put it into your calculator
does sin(7pi/6) = -1/2 and does sin(11pi/6) = -1/2
try it

- anonymous

I can use the unit circle too, yes?

- anonymous

if so, it is yes for both

- campbell_st

well there you go... sin can be negative... and the angles are larger then pi radians...

- anonymous

so thats the answer? no more steps?

- campbell_st

well as I said... that's my best guess...
and you told me the unit circle agreed...

- anonymous

so for 2... tan^2 = -3/2sec
(sin/cos )^2= -3/2 1/cos
??

- campbell_st

the unit circle produces the sin curve... that's important

- anonymous

trig is not my best section...

- 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

then multiply both sides by cos^2\[\sin^2(\theta) = \frac{-3 \cos(\theta)}{2}\]
is that ok..?

- anonymous

yes.. and then i square root everything?

- 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

okay.. so cross cancel the cos?

- 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

how do i set it up

- campbell_st

my advice is to use the general quadratic formula
\[\cos(\theta) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

- anonymous

but how since theres more than 1?

- 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

then only consider the value that is less than 1... a solution that is greater than 1 is undefined

- campbell_st

again, you need to consider where cos is positive... 1st and 4th quadrants...
so you will have 2 solutions...

- campbell_st

hope it made some sense... and good luck with your assessment task.

- anonymous

3.2+/- sqrt -3/2^2+4
-----
2
so 4.95 and 1.45?

- Jack1

how do u convert from radians to degrees again...?

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