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anonymous
 5 years ago
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
anonymous
 5 years ago
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i tried quadratic formula

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0ok...what do we have here....

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0quads fine.... what did you get for "u"? 2u^2 2u 1 = 0 where u = sin(x)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.02 + 2sqrt(2)  right? 4

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i probably messes that up in my head lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.04  (4)(2)(1) 12 = 4*3 2sqrt(3) is better

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0(1/2) + sqrt(3)/2 right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0sin(x) = (1/2) + sqrt(3)/2 sin(x) = sin(30) + cos(30) it looks like

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0and sin(x) = sin(30)  sin(60)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0sin...cos....keep em straight!! lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You need all degree solutions, yeah, not just over a fixed interval?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0sin(x) = 2cos(30) but i might have done it in error

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In 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
 5 years ago
Best ResponseYou've already chosen the best response.0338.51 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
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry, 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
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, amstre, you're on the right track.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry, about late response, but thank you for helping me
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