anonymous
  • anonymous
Help yet again with Trig! ( i tried it, but stuck at the end): find all possible degree solutions 2 sin ^2 x - 2 sin x -1=0
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
i tried quadratic formula
amistre64
  • amistre64
ok...what do we have here....
amistre64
  • amistre64
quads fine.... what did you get for "u"? 2u^2 -2u -1 = 0 where u = sin(x)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • amistre64
2 +- 2sqrt(2) ------------ right? 4
amistre64
  • amistre64
i probably messes that up in my head lol
amistre64
  • amistre64
4 - (4)(2)(-1) 12 = 4*3 2sqrt(3) is better
amistre64
  • amistre64
(1/2) +- sqrt(3)/2 right?
anonymous
  • anonymous
yes
amistre64
  • amistre64
sin(x) = (1/2) + sqrt(3)/2 sin(x) = sin(30) + cos(30) it looks like
amistre64
  • amistre64
and sin(x) = sin(30) - sin(60)
amistre64
  • amistre64
sin...cos....keep em straight!! lol
anonymous
  • anonymous
You need all degree solutions, yeah, not just over a fixed interval?
amistre64
  • amistre64
sin(x) = 2cos(30) but i might have done it in error
anonymous
  • anonymous
In general, \[\theta = n.180^o+(-1)^n \alpha^o\]where \[\alpha^o\]is a seed angle and\[n \in \mathbb{Z}\]You have the seed from the solution to the quadratic, \[\alpha^o = \sin^{-1}\left( \frac{1-\sqrt{3}}{2} \right)\]You have to reject the other solution since arcsine is not defined for an argument outside magnitude 1. So your solution should be,\[\theta =180^o+(-1)^n \sin^{-1}\left( \frac{1-\sqrt{3}}{2} \right)^o\] for n an integer.
amistre64
  • amistre64
338.51 degrees and 201.47 degrees are the 2 angles that would produce the proper sin(x) value.... right? then we add k2pi periods to those toget all the values....maybe :)
anonymous
  • anonymous
Sorry, in my last result, there's a typo. It should read:\[\theta = n.180^o+(-1)^n \sin ^ {-1} \left( \frac{1-\sqrt{3}}{2} \right)\]
anonymous
  • anonymous
Yeah, amstre, you're on the right track.
anonymous
  • anonymous
sorry, about late response, but thank you for helping me

Looking for something else?

Not the answer you are looking for? Search for more explanations.