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Math Help - unsure+ confused!?

  1. #1
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    unsure+ confused!?

    Solve 2sin^2 -7sin = -3 for , where 0°< < 360°.

    How would I do this???
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by superman View Post
    Solve 2sin^2 -7sin = -3 for , where 0°< < 360°.

    How would I do this???
    2 sin^2(theta) - 7 sin(theat) = -3

    is a quadratic in sin(theta). Solve this quadratic using to quadratic formula
    to get sin(theta) = 1/2 or 3. Now sin(theta)=3 has no solution for theta in
    the given range, so we are left with:

    sin(theta) = 1/2, for theta in (0, 360] degrees, which you should be able to do

    RonL
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  3. #3
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    Hello, superman!

    Solve: .2·sin²θ -7·sin θ .= .-3 .for 0°< θ < 360°.

    We have a quadratic equation: .2·sin²θ - 7·sin θ + 3 .= .0

    . . which factors: .(sin θ - 3)(2·sin θ - 1) .= .0


    And we have two equations to solve:

    . . sin θ - 3 .= .0 . . sin θ .= .3 . . . . no real solutions

    . . 2·sin θ - 1 .= .0 . . sin θ = ½ . . θ .= .30°, 150°

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