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animalsavior94

  • 3 years ago

Find all solutions in the interval [0, 2pi). give exact values and show all algebra. 2cos^2 X +3sinX-3=0

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  1. satellite73
    • 3 years ago
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    trick is to rewrite \[cos^2(x)= 1-sin^2(x)\]

  2. satellite73
    • 3 years ago
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    then get a quadratic equation in \[sin(x)\]

  3. satellite73
    • 3 years ago
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    \[2cos^2(x)+3sin(x)-3=0\] \[2(1-sin^2(x))+3sin(x)-3=0\] \[2-2sin^2(x)+3sin(x)-3=0\] \[-2sin^2(x)+3sin(x)-1=0\] \[2sin^2(x)-3sin(x)-1=0\]

  4. satellite73
    • 3 years ago
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    just solve as a quadratic. if it is easier to see replace sin(x) by z to get \[2z^2-3z+1=0\] \[(2z-1)(z-1)=0\] \[z=\frac{1}{2}\] or \[z=1\]

  5. satellite73
    • 3 years ago
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    then replace z by sin(x) and solve for x \[sin(x)=1\] \[x=\frac{\pi}{2}\]

  6. satellite73
    • 3 years ago
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    \[sin(x)=\frac{1}{2}\] \[x=\frac{\pi}{6}\] or \[x=\frac{5\pi}{6}\]

  7. satellite73
    • 3 years ago
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    ooops second answer is wrong sorry

  8. satellite73
    • 3 years ago
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    \[sin(x)=\frac{1}{2}\] \[x=\frac{11\pi}{6}\]

  9. satellite73
    • 3 years ago
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    k?

  10. animalsavior94
    • 3 years ago
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    so x=11pi/6?

  11. satellite73
    • 3 years ago
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    there are two places on the interval from 0 to 2pi where sine is 1/2

  12. satellite73
    • 3 years ago
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    one is at \[\frac{\pi}{6}\] and the other is at \[\frac{11\pi}{6}\] so all together you have 3 answers. those two and also \[\frac{\pi}{2}\] where sine is 1

  13. animalsavior94
    • 3 years ago
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    ohh i see thank you!

  14. satellite73
    • 3 years ago
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    welcome!

  15. satellite73
    • 3 years ago
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    any more?

  16. animalsavior94
    • 3 years ago
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    huh?

  17. satellite73
    • 3 years ago
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    any more of these to do?

  18. animalsavior94
    • 3 years ago
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    well i just put some up

  19. Gabysolis49
    • one year ago
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    Does anyone happen to know 2cos^2 x+sinx-1=0 solve for x on [0,2pi)

  20. Gabysolis49
    • one year ago
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    @satellite73

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