anonymous
  • anonymous
Please help! solve: 4sin³x+2sin²x-2sinx-1=0 for 0° ≤ x ≤ 360°
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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jdoe0001
  • jdoe0001
can you take common factor on the left-hand side?
anonymous
  • anonymous
common factor would be sin(x) right?
jdoe0001
  • jdoe0001
... h... actually... after getting rid of the 1, yes

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jdoe0001
  • jdoe0001
well... 2sin(x)
anonymous
  • anonymous
You add 1 to the right side but what would the left hand side look like after you simplify the gcf
jdoe0001
  • jdoe0001
\(\bf 4sin^3(x)+2sin^2(x)-2sin(x)-1=0\\ \quad \\ 4sin^3(x)+2sin^2(x)-2sin(x)=1\\ \quad \\ 2sin(x)\quad [2sin^2(x)+sin(x)-1] = 1\)
jdoe0001
  • jdoe0001
hmmmm
myininaya
  • myininaya
So are you looking to solve 4u^3+2u^2-2u-1=0 Did you try to find the possible 0's?
jdoe0001
  • jdoe0001
yea... I think so @myininaya is correct.. you may have to factor it with the 0 on the right-hand side
myininaya
  • myininaya
or you could factor by grouping :) much simpler
anonymous
  • anonymous
Yes I am trying to solve for the possible solutions in degrees. How would you factor by grouping?
myininaya
  • myininaya
\[(4\sin^3(x)+2\sin^2(x))+(-2\sin(x)-1)=0\]
myininaya
  • myininaya
Look at those first two terms. They have common factor. Look at the last terms. Factor out -1.
myininaya
  • myininaya
Or you can look at it in terms of u instead of sin at first. Your choice.
anonymous
  • anonymous
would it be 2sin^3(x) +sin(x)) + (2sin(x))=0 ?
myininaya
  • myininaya
\[(4u^3+2u^2)+(-2u-1)=0\]
myininaya
  • myininaya
What does 4u^3 and 2u^2 have in common?
anonymous
  • anonymous
2u^2? or just u^2?
myininaya
  • 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.
anonymous
  • anonymous
2u^2(2u+1) + ( 2sin) = 0
myininaya
  • myininaya
(-2u-1)=-(2u+1)
myininaya
  • myininaya
So (-2sin(x)-1)=-(2sin(x)+1)
myininaya
  • 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.
anonymous
  • anonymous
2u^2-1=0
myininaya
  • 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).
anonymous
  • anonymous
Oh yeah that's right, I haven't done factoring in awhile. what would be the next step after this one?
myininaya
  • myininaya
Setting both factors equal to 0.
anonymous
  • anonymous
sin x +1 = 0 sin x = -1 and sin^2x-1=0 sin^2x = 1
myininaya
  • myininaya
Your 2's are disappearing?
myininaya
  • myininaya
\[2u+1=0 \text{ or } 2u^2-1=0\] \[2u=-1 \text{ or } 2u^2=1\] \[u=\frac{-1}{2} \text{ or } u^2=\frac{1}{2}\]
anonymous
  • anonymous
How would you convert that to degrees for the final answer?
myininaya
  • myininaya
sin(x)=-1/2 Did I just show you the unit circle?
anonymous
  • anonymous
is it 210 and 330 degrees?
myininaya
  • 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
myininaya
  • myininaya
Now for this one, it maybe easy for you to rationalize the denominator first before looking on the unit circle.
anonymous
  • anonymous
so the four answers would be 30, 150, 210, and 330 degrees?
myininaya
  • myininaya
\[\sin(x)=\frac{\sqrt{2}}{2} \text{ or } \sin(x)=-\frac{\sqrt{2}}{2}\] I don't think you solved these correctly.
myininaya
  • myininaya
You solved sin(x)=-1/2 correctly. So part of your answer is 210 and 330 degrees.
anonymous
  • anonymous
okay thanks for the help
myininaya
  • myininaya
Np. You got it from here?
anonymous
  • anonymous
yeah i got it.
myininaya
  • myininaya
Neatness.

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