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

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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
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i tried quadratic formula
ok...what do we have here....
quads fine.... what did you get for "u"? 2u^2 -2u -1 = 0 where u = sin(x)

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2 +- 2sqrt(2) ------------ right? 4
i probably messes that up in my head lol
4 - (4)(2)(-1) 12 = 4*3 2sqrt(3) is better
(1/2) +- sqrt(3)/2 right?
yes
sin(x) = (1/2) + sqrt(3)/2 sin(x) = sin(30) + cos(30) it looks like
and sin(x) = sin(30) - sin(60)
sin...cos....keep em straight!! lol
You need all degree solutions, yeah, not just over a fixed interval?
sin(x) = 2cos(30) but i might have done it in error
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.
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 :)
Sorry, in my last result, there's a typo. It should read:\[\theta = n.180^o+(-1)^n \sin ^ {-1} \left( \frac{1-\sqrt{3}}{2} \right)\]
Yeah, amstre, you're on the right track.
sorry, about late response, but thank you for helping me

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