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anonymous

  • one year ago

Find all solutions in the interval [0, 2π). sin2 x + sin x = 0

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  1. anonymous
    • one year ago
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    i give medals! please help!

  2. anonymous
    • one year ago
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    you can factor the common term of sin x sin² x + sin x = 0 (sin x)(sin x + 1) = 0 Now set both factors equal to 0 and solve these two equations sin x = 0 sin x + 1 = 0

  3. anonymous
    • one year ago
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    I am very confused

  4. anonymous
    • one year ago
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    i only have a few more minutes to answer this please help and explain!

  5. Australopithecus
    • one year ago
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    So you have, \[\sin^2(x) + \sin(x) = 0\] By the definition of a exponent \[x^2 = x*x\] etc so, \[\sin^2(x) + \sin(x) = 0\] can be written as: \[\sin(x)\sin(x) + \sin(x) = 0\] Now you can factor out sin(x): \[\sin(x)(\sin(x) + 1) = 0\] Now look at both terms sin(x) and (1 + sin(x)) When one of them equals 0 you have a solution

  6. anonymous
    • one year ago
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    sin² x and sin x have a common term of sin x so it can be factored out giving the equation as (sin x)(sin x + 1) = 0 If two thing multiply to be 0, one or both of them is 0. so we can write thes the two equations sin x = 0 and sin x + 1 = 0 To solve sin x = 0, look on the unit circle and pick out all the angle where sine is 0. To solve sin x + 1 = 0, subtract the 1 from both sides sin x = -1 now pick out the angles that have a sine of -1.

  7. Australopithecus
    • one year ago
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    so for (1 + sin(x)) = 0 sin(x) must equal - 1 what value of x makes sin(x) = -1? For the second term sin(x) = 0 what value of x makes sin(x) equal 0 Look at the graph of sin(x) http://www.wolframalpha.com/input/?i=graph+of+sin%28x%29+between+0+and+2pi

  8. Australopithecus
    • one year ago
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    Remember that you are only looking for solutions for the interval [0,2pi)

  9. anonymous
    • one year ago
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    is there answer x= 0,pi,4pi/3,5pi/3?

  10. anonymous
    • one year ago
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    0 and pi are solutions to sin x = 0. the other two aren't solutions to either equation. Where is sine equal to -1 between 0 and 2pi?

  11. anonymous
    • one year ago
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    x=o,pi,pi/3,2pi/3?

  12. Australopithecus
    • one year ago
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    when x = 0 sin(0) = 0 so x =0 is a solution sin(4pi/3) does not equal 0 or -1 so it is not a solution sin(5pi/3) does not equal 0 or -1 so it is not a solution sin(pi) = 0 so it is a solution

  13. Australopithecus
    • one year ago
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    Look at the interval 0 to 2pi It is just the top section of the unit circle

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  14. anonymous
    • one year ago
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    0, pi, 3pi/2

  15. Australopithecus
    • one year ago
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    yup

  16. anonymous
    • one year ago
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    Thanks!

  17. Australopithecus
    • one year ago
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    If you found my help useful please rate me using the qualified help located at the top

  18. Australopithecus
    • one year ago
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    So I can get paid :)

  19. anonymous
    • one year ago
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    I will!

  20. Australopithecus
    • one year ago
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    Not that I wouldnt have helped you for free :)

  21. anonymous
    • one year ago
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    How do you do that?

  22. Australopithecus
    • one year ago
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    http://i.imgur.com/7yfPY5A.jpg

  23. Australopithecus
    • one year ago
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    Click this button :)

  24. Australopithecus
    • one year ago
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    Ha brain is in stupid mode right now Click rate qualified helper so the button next to it

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