1. jdoe0001

can you take common factor on the left-hand side?

2. baseballguy1322

common factor would be sin(x) right?

3. jdoe0001

... h... actually... after getting rid of the 1, yes

4. jdoe0001

well... 2sin(x)

5. baseballguy1322

You add 1 to the right side but what would the left hand side look like after you simplify the gcf

6. jdoe0001

$$\bf 4sin^3(x)+2sin^2(x)-2sin(x)-1=0\\ \quad \\ 4sin^3(x)+2sin^2(x)-2sin(x)=1\\ \quad \\ 2sin(x)\quad [2sin^2(x)+sin(x)-1] = 1$$

7. jdoe0001

hmmmm

8. myininaya

So are you looking to solve 4u^3+2u^2-2u-1=0 Did you try to find the possible 0's?

9. jdoe0001

yea... I think so @myininaya is correct.. you may have to factor it with the 0 on the right-hand side

10. myininaya

or you could factor by grouping :) much simpler

11. baseballguy1322

Yes I am trying to solve for the possible solutions in degrees. How would you factor by grouping?

12. myininaya

$(4\sin^3(x)+2\sin^2(x))+(-2\sin(x)-1)=0$

13. myininaya

Look at those first two terms. They have common factor. Look at the last terms. Factor out -1.

14. myininaya

Or you can look at it in terms of u instead of sin at first. Your choice.

15. baseballguy1322

would it be 2sin^3(x) +sin(x)) + (2sin(x))=0 ?

16. myininaya

$(4u^3+2u^2)+(-2u-1)=0$

17. myininaya

What does 4u^3 and 2u^2 have in common?

18. baseballguy1322

2u^2? or just u^2?

19. myininaya

The gcf would be 2u^2 So we have $2u^2(2u+1)+(-2u-1)=0$ Those last two terms, you can factor out -1.

20. baseballguy1322

2u^2(2u+1) + ( 2sin) = 0

21. myininaya

(-2u-1)=-(2u+1)

22. myininaya

So (-2sin(x)-1)=-(2sin(x)+1)

23. myininaya

$2u^2(2u+1)-1(2u+1)=0$ Now you have two terms. (2u+1) is a factor of both of those terms. Do you think you can completely factor the left hand side now.

24. baseballguy1322

2u^2-1=0

25. myininaya

Don't forget about the (2u+1) So to factor $2u^2(2u+1)-1(2u+1) \text{ we find what both of our terms have in common }$ The (2u+1) So we factor the (2u+1) out giving us $(2u+1)(2u^2-1)=0$ Set both factors equal to 0. Don't forget u is sin(x).

26. baseballguy1322

Oh yeah that's right, I haven't done factoring in awhile. what would be the next step after this one?

27. myininaya

Setting both factors equal to 0.

28. baseballguy1322

sin x +1 = 0 sin x = -1 and sin^2x-1=0 sin^2x = 1

29. myininaya

30. myininaya

$2u+1=0 \text{ or } 2u^2-1=0$ $2u=-1 \text{ or } 2u^2=1$ $u=\frac{-1}{2} \text{ or } u^2=\frac{1}{2}$

31. baseballguy1322

How would you convert that to degrees for the final answer?

32. myininaya

sin(x)=-1/2 Did I just show you the unit circle?

33. baseballguy1322

is it 210 and 330 degrees?

34. myininaya

You also need to solve sin^2(x)=1/2 Take the square root of both sides $\sin(x)=\pm \sqrt{ \frac{1}{2} }$ Don't forget to solve these two as well. And yes those are the solutions to sin(x)=-1/2

35. myininaya

Now for this one, it maybe easy for you to rationalize the denominator first before looking on the unit circle.

36. baseballguy1322

so the four answers would be 30, 150, 210, and 330 degrees?

37. myininaya

$\sin(x)=\frac{\sqrt{2}}{2} \text{ or } \sin(x)=-\frac{\sqrt{2}}{2}$ I don't think you solved these correctly.

38. myininaya

You solved sin(x)=-1/2 correctly. So part of your answer is 210 and 330 degrees.

39. baseballguy1322

okay thanks for the help

40. myininaya

Np. You got it from here?

41. baseballguy1322

yeah i got it.

42. myininaya

Neatness.