kittiwitti1
  • kittiwitti1
http://prntscr.com/9auym5 Tried using quadratic formula and got a nonreal answer, what the heck o-0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
kittiwitti1
  • kittiwitti1
Okay, yes; definitely a nonreal answer?! http://prntscr.com/9auzc7
jim_thompson5910
  • jim_thompson5910
graph the left side as a function https://www.desmos.com/calculator then see where it crosses the x axis
jim_thompson5910
  • jim_thompson5910
you can click on the roots to have the coordinates show up you'll have to zoom in to get more decimal places

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kittiwitti1
  • kittiwitti1
http://prntscr.com/9auzw8 I think this website hates me.
jim_thompson5910
  • jim_thompson5910
you have to type in `(sin(x))^2` instead of sin^2(x)
kittiwitti1
  • kittiwitti1
ah
kittiwitti1
  • kittiwitti1
http://prntscr.com/9av0c3 0-0
jim_thompson5910
  • jim_thompson5910
don't enter the `=0` part just the left hand side
kittiwitti1
  • kittiwitti1
oh, okay
kittiwitti1
  • kittiwitti1
I got the equation in, what should I do now lol
kittiwitti1
  • kittiwitti1
I never used this site before, but it seems pretty nifty. Thanks for the reference :)
jim_thompson5910
  • jim_thompson5910
click on the roots to have the coordinates show up roots = x intercepts
kittiwitti1
  • kittiwitti1
Okay
kittiwitti1
  • kittiwitti1
like this? http://prntscr.com/9av0wy
jim_thompson5910
  • jim_thompson5910
wherever the graph of f(x) crosses the x axis is going to represent an approximate solution to f(x) = 0
jim_thompson5910
  • jim_thompson5910
yeah but keep in mind that 0 <= x < 2pi
kittiwitti1
  • kittiwitti1
http://prntscr.com/9av12t
jim_thompson5910
  • jim_thompson5910
still not in the right interval
kittiwitti1
  • kittiwitti1
eh o_o
kittiwitti1
  • kittiwitti1
I'm lost then xD
jim_thompson5910
  • jim_thompson5910
1 Attachment
kittiwitti1
  • kittiwitti1
No, I get that; but I'm not sure how to apply the limits to the graph
jim_thompson5910
  • jim_thompson5910
x has to be both larger than 0 AND smaller than 2pi = 6.28 approx
kittiwitti1
  • kittiwitti1
ah
jim_thompson5910
  • jim_thompson5910
you can click and drag to move the window of the graph
kittiwitti1
  • kittiwitti1
http://prntscr.com/9av1mr these?
jim_thompson5910
  • jim_thompson5910
yep now just zoom in until it shows you 4 decimal places
kittiwitti1
  • kittiwitti1
alright!
kittiwitti1
  • kittiwitti1
http://prntscr.com/9av1wa Wow, this is really helpful, thanks @jim_thompson5910
jim_thompson5910
  • jim_thompson5910
you nailed them both
DanJS
  • DanJS
if sin(x) = u u^2 - 3u - 3 = 0 \[u = \frac{ 3 \pm \sqrt{9-4*1*(-3)} }{ 2 }\] one works
DanJS
  • DanJS
sin(x) = u sin^(-1)(u) = x
kittiwitti1
  • kittiwitti1
._.;
jim_thompson5910
  • jim_thompson5910
@DanJS is showing how to find the solutions without graphing
kittiwitti1
  • kittiwitti1
Ah. It asked for graphing though ^^;
DanJS
  • DanJS
the negative one , subtracting the root, will give you a good answer in the 0 to 2pi interval
kittiwitti1
  • kittiwitti1
O_O okay
kittiwitti1
  • kittiwitti1
Well I'd give you a medal but I already gave one to jim ;-; I'm sorry!
DanJS
  • DanJS
inverse sin of that , you get something like 0.91291 the other is a neg angle
DanJS
  • DanJS
not real
DanJS
  • DanJS
ha, thats fine,
kittiwitti1
  • kittiwitti1
>.<;;
kittiwitti1
  • kittiwitti1
But thank you for the info :D

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