will give medals!
can some one tell me where i go from here in this trig problem? cos 2θ + 5 cos θ = 6
i posted what ive done so far, below:

- x3_drummerchick

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- x3_drummerchick

|dw:1449935885909:dw|

- zepdrix

\(\large\rm u+w=\theta\) ?

- x3_drummerchick

i know im supposed to do cosθ = -7/2 and cosθ=1, and evaluate, but i thought -7/2 is out of cosines domain

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## More answers

- x3_drummerchick

let w = θ

- x3_drummerchick

my writing is just messy

- zepdrix

Oh :)
Then I think you meant let w=cos(theta)
Beside the point though hehe

- mathmale

Good thing you're sharing what you've already done. I was about to interpret your cos 2x as "the cosine of twice the angle x," but apparently you meant [cos x]^2.
If you mean [cos x]^2, write it as |dw:1449936128460:dw|

- mathmale

or as\[\cos^2x\]

- x3_drummerchick

yes, sorry i was rushing, i understand the subtitution part, as shown above in my comment ^

- mathmale

but definitely not as cos 2x.

- x3_drummerchick

one sec, heres a neater version:

- x3_drummerchick

##### 1 Attachment

- zepdrix

He did in fact mean cos(2x).
He applied Double Angle in the next step.

- mathmale

If you meant to present\[\cos^2\theta+5\cos \theta=6, \]

- zepdrix

She she she* sorry :3

- x3_drummerchick

yes, i attached a picture to show what ive done ^, my writing on here is awful , forgive me

- mathmale

you could model this using x as your variable:\[x^2+5x-6=0\]

- mathmale

Can you solve this for x? If so, do exactly the same thing when working with cos theta instead of x.

- x3_drummerchick

i did, i already got the values for the x, i finished up with cosθ being -7/2 and 1. im not sure what to do from there to get the exact values

- x3_drummerchick

the picture i posted shows my work....

- mathmale

All right. I thanked you earlier for having shown your work.

- zepdrix

Your work looks good drum :)
And yes, -7/2 is probably extraneous in this case.

- mathmale

How would you know whether or not -7/2 is extraneous?

- x3_drummerchick

so then the other value cosθ= 1 is just 0, or pi/2 + 2pk?

- x3_drummerchick

iits out of cosines domain

- zepdrix

just 0, yah?
Not sure where the pi/2 is coming from :))

- zepdrix

0+2k pi

- x3_drummerchick

-1<=x<=1 (domain of cosine)

- x3_drummerchick

i thought it had to be in radians? lol you have to take the arc cos of both sides

- x3_drummerchick

|dw:1449936549380:dw|

- zepdrix

\[\large\rm \cos \theta=1\qquad\to\qquad \arccos(1)=\theta\]

- zepdrix

\[\large\rm \theta=0+2k \pi\]

- zepdrix

pi/2 would be the solution angle for sin(theta)=1

- x3_drummerchick

or just 2(pi)k? can i write that as well?

- zepdrix

Ah, yes :)
That probably better than leaving that ugly zero there lol

- x3_drummerchick

thank you :)
the other helper wasnt so helpful. he was reiterating what i already knew :p

- zepdrix

Here is the graph of the function just in case you are curious,
https://www.desmos.com/calculator/idqfjgw4m1
always nice when you can verify your work.
Notice it only touches the x-axis at multiples of 2pi.

- x3_drummerchick

oh wow! i didnt know that

- zepdrix

Just in case you're wondering how I did that:
You subtract 6 to the other side, ya?\[\large\rm \cos2\theta+5\cos \theta-6=0\]And then just graph\[\large\rm \cos2\theta+5\cos \theta-6=y\]And look for the places y is at a value of 0.

- zepdrix

That's why I was looking at the x-axis.
Hopefully that's clear.

- x3_drummerchick

that makes sense. Thank you :)

- zepdrix

np c:

- x3_drummerchick

poop. it was just 0. x( i thought he wanted radians

- zepdrix

Keep in mind that 0 degrees = 0 radians.
Was a specific interval included maybe?
Like "Find solutions on the interval [0,2pi)"?

- x3_drummerchick

yes

- zepdrix

Ah you silly billy,
then radians is fine,
but they only wanted solutions within one full rotation of the unit circle.
Only the angles between 0 and 2pi, excluding the 2pi end point.
>.<

- x3_drummerchick

its all good, ima ace this next one :p its just our final review assignment

- zepdrix

ah :)

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