## baseballguy1322 Group Title Please help! solve: 4sin³x+2sin²x-2sinx-1=0 for 0° ≤ x ≤ 360° 9 months ago 9 months ago

1. jdoe0001 Group Title

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

2. baseballguy1322 Group Title

common factor would be sin(x) right?

3. jdoe0001 Group Title

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

4. jdoe0001 Group Title

well... 2sin(x)

5. baseballguy1322 Group Title

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

6. jdoe0001 Group Title

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

7. jdoe0001 Group Title

hmmmm

8. myininaya Group Title

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

9. jdoe0001 Group Title

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

10. myininaya Group Title

or you could factor by grouping :) much simpler

11. baseballguy1322 Group Title

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

12. myininaya Group Title

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

13. myininaya Group Title

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

14. myininaya Group Title

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

15. baseballguy1322 Group Title

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

16. myininaya Group Title

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

17. myininaya Group Title

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

18. baseballguy1322 Group Title

2u^2? or just u^2?

19. myininaya Group Title

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 Group Title

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

21. myininaya Group Title

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

22. myininaya Group Title

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

23. myininaya Group Title

$2u^2(2u+1)-1(2u+1)=0$ 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 Group Title

2u^2-1=0

25. myininaya Group Title

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 Group Title

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

27. myininaya Group Title

Setting both factors equal to 0.

28. baseballguy1322 Group Title

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

29. myininaya Group Title

Your 2's are disappearing?

30. myininaya Group Title

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

31. baseballguy1322 Group Title

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

32. myininaya Group Title

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

33. baseballguy1322 Group Title

is it 210 and 330 degrees?

34. myininaya Group Title

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 Group Title

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

36. baseballguy1322 Group Title

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

37. myininaya Group Title

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

38. myininaya Group Title

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

39. baseballguy1322 Group Title

okay thanks for the help

40. myininaya Group Title

Np. You got it from here?

41. baseballguy1322 Group Title

yeah i got it.

42. myininaya Group Title

Neatness.