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Homework Help: Find all solutions of the equation 3sin^2x-7sinx+2=0

  1. Nov 16, 2009 #1
    Find all solutions of the equation 3sin^2x-7sinx+2=0 in the interval [0,2pi).

    I factored it out first and got (3sinx-1)(sinx+2)=0 and then solved, and got sinx=1/3 and sinx=-2 if I did it correctly. What i am having trouble with is finding where sin(x)=1/3 and -2.
  2. jcsd
  3. Nov 16, 2009 #2


    Staff: Mentor

    One of your factors is incorrect.

    In your interval [0, 2pi) there are two solutions to sin x = 1/3. Your calculator will give you one of them, and you'll need an identity to figure out the other one.

    To solve sin x = - 2 (which is incorrect for this problem), first think about the range of the sine function.
  4. Nov 16, 2009 #3
    You factored incorrectly. That should be (3sinx-1)(sinx-2)=0

    So sin x = 1/3 and 2

    Take the sine inverse of the two.
    Sin x = 2. csc 30 = 2 so sin- (1/2) = x. x=30 degrees or pi/6 rad.
    sin x = 1/3. sin- (1/3)=x. x=19.47 degrees or .34pi.

    That's what I would've done but I'm not sure if that was correct or not, so don't base your work on my steps.

    If it were correct, one should be able to find the solutions to the equation from there. Recall that sine is only positive in quadrant 1 and 2.
    Last edited: Nov 16, 2009
  5. Nov 16, 2009 #4


    Staff: Mentor

    So are you saying here that sin(30 deg.) = 2?
    The measure of the angle is not .34 pi. That would be a little over 1 radian, which is way bigger than 19+ degrees.
  6. Nov 16, 2009 #5
    Sorry, I didn't mean .34 pi. I just meant .34 rad.

    And as for the sin(30) = 2, likely no. That's why I said to him to not base his work on my steps. I was hoping someone would correct it and I'd learn something new. I just recall something about cot x = a. tan x = (1/a) x = tan- (1/a). I guess that's where I derived it from but yes, it would still be incorrect and that is why I said what I said. But come to think of it, that wouldn't be possible as the opposite side can not be longer than the hypotenuse.
    Last edited: Nov 16, 2009
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