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dragonsly

  • 2 years ago

how would you find the solution for this sin^2x - .49=0<--- the answer has 4 different solutions

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  1. Algebraic!
    • 2 years ago
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    (sinx)^2 =.49 sin(x) = +/- .49 doesn't give an interval, so it has infinite solutions... if your interval is something like 0..2pi then arcsin(-.49) (two solutions) and arcsin(.49) (two more solutions)

  2. dragonsly
    • 2 years ago
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    the intervals are between 0 and 2pie

  3. dragonsly
    • 2 years ago
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    and wouldn't it be root .49 because sin is squrd?

  4. Algebraic!
    • 2 years ago
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    good call...

  5. Algebraic!
    • 2 years ago
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    (sinx)^2 =.49 sin(x) = +/- sqrt(.49) doesn't give an interval, so it has infinite solutions... if your interval is something like 0..2pi then arcsin(-sqrt(.49) (two solutions) and arcsin(sqrt(.49) ) (two more solutions)

  6. dragonsly
    • 2 years ago
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    but how would you find the solution would i add 3.14 to root .49 and subtract 3.14 (btw everythings in radians)

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