baseballguy1322
Please help! solve: 4sin³x+2sin²x-2sinx-1=0 for 0° ≤ x ≤ 360°
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jdoe0001
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can you take common factor on the left-hand side?
baseballguy1322
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common factor would be sin(x) right?
jdoe0001
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... h... actually... after getting rid of the 1, yes
jdoe0001
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well... 2sin(x)
baseballguy1322
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You add 1 to the right side but what would the left hand side look like after you simplify the gcf
jdoe0001
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\(\bf 4sin^3(x)+2sin^2(x)-2sin(x)-1=0\\ \quad \\
4sin^3(x)+2sin^2(x)-2sin(x)=1\\ \quad \\
2sin(x)\quad [2sin^2(x)+sin(x)-1] = 1\)
jdoe0001
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hmmmm
myininaya
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So are you looking to solve 4u^3+2u^2-2u-1=0
Did you try to find the possible 0's?
jdoe0001
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yea... I think so @myininaya is correct.. you may have to factor it with the 0 on the right-hand side
myininaya
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or you could factor by grouping :) much simpler
baseballguy1322
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Yes I am trying to solve for the possible solutions in degrees. How would you factor by grouping?
myininaya
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\[(4\sin^3(x)+2\sin^2(x))+(-2\sin(x)-1)=0\]
myininaya
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Look at those first two terms. They have common factor.
Look at the last terms. Factor out -1.
myininaya
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Or you can look at it in terms of u instead of sin at first. Your choice.
baseballguy1322
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would it be 2sin^3(x) +sin(x)) + (2sin(x))=0 ?
myininaya
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\[(4u^3+2u^2)+(-2u-1)=0\]
myininaya
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What does 4u^3 and 2u^2 have in common?
baseballguy1322
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2u^2? or just u^2?
myininaya
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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.
baseballguy1322
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2u^2(2u+1) + ( 2sin) = 0
myininaya
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(-2u-1)=-(2u+1)
myininaya
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So (-2sin(x)-1)=-(2sin(x)+1)
myininaya
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\[2u^2(2u+1)-1(2u+1)=0\]
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.
baseballguy1322
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2u^2-1=0
myininaya
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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).
baseballguy1322
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Oh yeah that's right, I haven't done factoring in awhile. what would be the next step after this one?
myininaya
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Setting both factors equal to 0.
baseballguy1322
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sin x +1 = 0
sin x = -1
and
sin^2x-1=0
sin^2x = 1
myininaya
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Your 2's are disappearing?
myininaya
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\[2u+1=0 \text{ or } 2u^2-1=0\]
\[2u=-1 \text{ or } 2u^2=1\]
\[u=\frac{-1}{2} \text{ or } u^2=\frac{1}{2}\]
baseballguy1322
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How would you convert that to degrees for the final answer?
myininaya
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sin(x)=-1/2
Did I just show you the unit circle?
baseballguy1322
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is it 210 and 330 degrees?
myininaya
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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
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Now for this one, it maybe easy for you to rationalize the denominator first before looking on the unit circle.
baseballguy1322
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so the four answers would be 30, 150, 210, and 330 degrees?
myininaya
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\[\sin(x)=\frac{\sqrt{2}}{2} \text{ or } \sin(x)=-\frac{\sqrt{2}}{2}\]
I don't think you solved these correctly.
myininaya
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You solved sin(x)=-1/2 correctly.
So part of your answer is 210 and 330 degrees.
baseballguy1322
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okay thanks for the help
myininaya
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Np. You got it from here?
baseballguy1322
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yeah i got it.
myininaya
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Neatness.