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sabika13

  • 3 years ago

Find the points of intersection of each pair of curves in the given interval. i) y=sin2x, y=sinx

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  1. dpaInc
    • 3 years ago
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    set the y's equal...: sin2x = sinx given 2sinxcosx = sinx use the double angle formula for sine 2sinxcosx - sinx = 0 move the sinx over to the left side sinx(2cosx - 1) = 0 factor can you do the rest from here?

  2. dpaInc
    • 3 years ago
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    and btw, what is the interval?

  3. surdawi
    • 3 years ago
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    \[(0 \pm 2*\pi*n,0),(\pi \pm 2*\pi*n,0)\]

  4. sabika13
    • 3 years ago
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    I got 0,0 but theres more intervals that icant get

  5. dpaInc
    • 3 years ago
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    no... what interval do you want to look for solution(s)? as it is stated in the problem, "... in the given interval"

  6. dpaInc
    • 3 years ago
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    what is the given interval?

  7. sabika13
    • 3 years ago
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    SOrry the given interval is 0</= x </= 2pi

  8. dpaInc
    • 3 years ago
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    ok... the interval is [0, 2pi]

  9. dpaInc
    • 3 years ago
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    so from the last equation: sinx(2cosx - 1) = 0 you'll need to set each factor equal to zero... sinx = 0 solve this... and also 2cosx - 1 = 0 solve this....

  10. dpaInc
    • 3 years ago
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    in the first equation, what angle between 0 and 2pi will give you a sine of zero?

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