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anonymous

  • 4 years ago

Solve 4sin(2y - 0.3) + 5cos(27 - 0.3) = 0, for 0(lesser-than-or-equal-to) y (lesser-than-or-equal-to) "pi" radians.

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  1. anonymous
    • 4 years ago
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    Tanvi is that 5cos(27-0.3)

  2. anonymous
    • 4 years ago
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    sorry, 2y - 0.3, not 27 -_-

  3. anonymous
    • 4 years ago
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    hhmm - i dont recall solving one like this before - i'll have to check my list of trig identities - wait a few minutes

  4. anonymous
    • 4 years ago
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    sure, suree, take your time. even if we replace (2y - 0.3) by a, then we would get 4sin(a) + 5cos(a) = 0. oh damn, i got it!!

  5. anonymous
    • 4 years ago
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    tats the way to go

  6. anonymous
    • 4 years ago
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    i don't know what that means, but it'll become tan (a) = -5/4, and then i'll solve it.

  7. anonymous
    • 4 years ago
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    yes - thats an easier way - do you really need anyone to help you tanvidals??!! lol well done

  8. anonymous
    • 4 years ago
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    yes, can you give me another value for a??? apart from -0.09 rad?? it needs to be positive.

  9. anonymous
    • 4 years ago
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    tan -5/4 - the possible values are between o and 2pi radians are in the second and 4th quadrant so we get pi - 0.9 and 2pi - 0.9

  10. anonymous
    • 4 years ago
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    i need between 0 and pi.

  11. anonymous
    • 4 years ago
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    yes so a = 2.24 , 5.38 so 2y - 0.3 = 2.24 y = 2.54 / 2 = 1.27 rad a=5.38 will be too big one value for y = 1.27 rad

  12. anonymous
    • 4 years ago
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    i needed to go higher than pi for value of a because y must have value 0 to pi

  13. anonymous
    • 4 years ago
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    thank youu :)

  14. anonymous
    • 4 years ago
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    yw - but i think ill have another coffee to stimulate my brain a bit!

  15. anonymous
    • 4 years ago
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    oh - plz click on good answer to close the question if you are happy with the answer

  16. anonymous
    • 4 years ago
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    i've just checked the answer on wolfram alpha - it gives it as 1.2727

  17. anonymous
    • 4 years ago
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    http://www.wolframalpha.com/input/?i=Solve+4sin%282y+-+0.3%29+%2B+5cos%282y+-+0.3%29+%3D+0

  18. anonymous
    • 4 years ago
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    thats a good site for verification

  19. anonymous
    • 4 years ago
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    run your cursor over the root plot - the red spot on positive x-axis - i gives you this value

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