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Please help! solve: 4sin³x+2sin²x-2sinx-1=0 for 0° ≤ x ≤ 360°

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can you take common factor on the left-hand side?
common factor would be sin(x) right?
... h... actually... after getting rid of the 1, yes

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Other answers:

well... 2sin(x)
You add 1 to the right side but what would the left hand side look like after you simplify the gcf
\(\bf 4sin^3(x)+2sin^2(x)-2sin(x)-1=0\\ \quad \\ 4sin^3(x)+2sin^2(x)-2sin(x)=1\\ \quad \\ 2sin(x)\quad [2sin^2(x)+sin(x)-1] = 1\)
So are you looking to solve 4u^3+2u^2-2u-1=0 Did you try to find the possible 0's?
yea... I think so @myininaya is correct.. you may have to factor it with the 0 on the right-hand side
or you could factor by grouping :) much simpler
Yes I am trying to solve for the possible solutions in degrees. How would you factor by grouping?
Look at those first two terms. They have common factor. Look at the last terms. Factor out -1.
Or you can look at it in terms of u instead of sin at first. Your choice.
would it be 2sin^3(x) +sin(x)) + (2sin(x))=0 ?
What does 4u^3 and 2u^2 have in common?
2u^2? or just u^2?
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.
2u^2(2u+1) + ( 2sin) = 0
So (-2sin(x)-1)=-(2sin(x)+1)
\[2u^2(2u+1)-1(2u+1)=0\] 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.
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).
Oh yeah that's right, I haven't done factoring in awhile. what would be the next step after this one?
Setting both factors equal to 0.
sin x +1 = 0 sin x = -1 and sin^2x-1=0 sin^2x = 1
Your 2's are disappearing?
\[2u+1=0 \text{ or } 2u^2-1=0\] \[2u=-1 \text{ or } 2u^2=1\] \[u=\frac{-1}{2} \text{ or } u^2=\frac{1}{2}\]
How would you convert that to degrees for the final answer?
sin(x)=-1/2 Did I just show you the unit circle?
is it 210 and 330 degrees?
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
Now for this one, it maybe easy for you to rationalize the denominator first before looking on the unit circle.
so the four answers would be 30, 150, 210, and 330 degrees?
\[\sin(x)=\frac{\sqrt{2}}{2} \text{ or } \sin(x)=-\frac{\sqrt{2}}{2}\] I don't think you solved these correctly.
You solved sin(x)=-1/2 correctly. So part of your answer is 210 and 330 degrees.
okay thanks for the help
Np. You got it from here?
yeah i got it.

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