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anonymous

  • 5 years ago

find the exact solutions of the given equation in the interval [0,2pi] cos2x+3cosx+2=0

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  1. myininaya
    • 5 years ago
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    \[\cos^2(x)+3\cos(x)+2=0?\]

  2. anonymous
    • 5 years ago
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    it is cos(2x) +3cos(x)+2=0

  3. anonymous
    • 5 years ago
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    \[(z+1)(z+2)=0\] \[z=-1\] \[z=-2\] \[\cos(x)=-1\] \[\cos(x)=-2\] second one is very unlikely

  4. anonymous
    • 5 years ago
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    im sorry im not sure i follow. they have to be in the interval [0,2pi]

  5. anonymous
    • 5 years ago
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    cos (2x) = 2 cos^2 x - 1 2 cos^2 x + 3 cos x + 1 = 0

  6. myininaya
    • 5 years ago
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    \[\cos(2x)=\cos^2(x)-\sin^2(x)=\cos^2(x)-(1-\cos^2(x))=2\cos^2(x)-1\]

  7. myininaya
    • 5 years ago
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    \[2\cos^2(x)-1+3\cos(x)+2=0\] \[2\cos^2(x)+3\cos(x)+2-1=0\] \[2\cos^2(x)+3\cos(x)+1=0\] Can you factor? \[2u^2+3u+1=0\] (note the relationship is that u=cos(x))

  8. myininaya
    • 5 years ago
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    \[2u^2+2u+1u+1=0\] since 2u+1u=3u now we factor by grouping \[2u(u+1)+1(u+1)=0\] \[(u+1)(2u+1)=0\]

  9. myininaya
    • 5 years ago
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    =>\[u=-1 \text{ or } u=\frac{-1}{2}\]

  10. myininaya
    • 5 years ago
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    but remember u =cos(x)

  11. myininaya
    • 5 years ago
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    \[\cos(x)=-1 \text{ or } \cos(x)=\frac{-1}{2}\]

  12. anonymous
    • 5 years ago
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    yes i got it now! thank you :)))

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