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\(\large\rm u+w=\theta\) ?
i know im supposed to do cosθ = -7/2 and cosθ=1, and evaluate, but i thought -7/2 is out of cosines domain
let w = θ
my writing is just messy
Oh :) Then I think you meant let w=cos(theta) Beside the point though hehe
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|
yes, sorry i was rushing, i understand the subtitution part, as shown above in my comment ^
but definitely not as cos 2x.
one sec, heres a neater version:
He did in fact mean cos(2x). He applied Double Angle in the next step.
If you meant to present\[\cos^2\theta+5\cos \theta=6, \]
She she she* sorry :3
yes, i attached a picture to show what ive done ^, my writing on here is awful , forgive me
you could model this using x as your variable:\[x^2+5x-6=0\]
Can you solve this for x? If so, do exactly the same thing when working with cos theta instead of x.
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
the picture i posted shows my work....
All right. I thanked you earlier for having shown your work.
Your work looks good drum :) And yes, -7/2 is probably extraneous in this case.
How would you know whether or not -7/2 is extraneous?
so then the other value cosθ= 1 is just 0, or pi/2 + 2pk?
iits out of cosines domain
just 0, yah? Not sure where the pi/2 is coming from :))
-1<=x<=1 (domain of cosine)
i thought it had to be in radians? lol you have to take the arc cos of both sides
\[\large\rm \cos \theta=1\qquad\to\qquad \arccos(1)=\theta\]
\[\large\rm \theta=0+2k \pi\]
pi/2 would be the solution angle for sin(theta)=1
or just 2(pi)k? can i write that as well?
Ah, yes :) That probably better than leaving that ugly zero there lol
thank you :) the other helper wasnt so helpful. he was reiterating what i already knew :p
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.
oh wow! i didnt know that
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.
That's why I was looking at the x-axis. Hopefully that's clear.
that makes sense. Thank you :)
poop. it was just 0. x( i thought he wanted radians
Keep in mind that 0 degrees = 0 radians. Was a specific interval included maybe? Like "Find solutions on the interval [0,2pi)"?
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. >.<
its all good, ima ace this next one :p its just our final review assignment