## 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

1. anonymous

2. amistre64

ok...what do we have here....

3. amistre64

quads fine.... what did you get for "u"? 2u^2 -2u -1 = 0 where u = sin(x)

4. amistre64

2 +- 2sqrt(2) ------------ right? 4

5. amistre64

i probably messes that up in my head lol

6. amistre64

4 - (4)(2)(-1) 12 = 4*3 2sqrt(3) is better

7. amistre64

(1/2) +- sqrt(3)/2 right?

8. anonymous

yes

9. amistre64

sin(x) = (1/2) + sqrt(3)/2 sin(x) = sin(30) + cos(30) it looks like

10. amistre64

and sin(x) = sin(30) - sin(60)

11. amistre64

sin...cos....keep em straight!! lol

12. anonymous

You need all degree solutions, yeah, not just over a fixed interval?

13. amistre64

sin(x) = 2cos(30) but i might have done it in error

14. 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.

15. 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 :)

16. anonymous

Sorry, in my last result, there's a typo. It should read:$\theta = n.180^o+(-1)^n \sin ^ {-1} \left( \frac{1-\sqrt{3}}{2} \right)$

17. anonymous

Yeah, amstre, you're on the right track.

18. anonymous

sorry, about late response, but thank you for helping me