# Math 12

posted by .

trig question
2sin^2x+sinx-6=0
how to solve for x when the
restriction is x is greater than or equal to zero but less than 2pi

• Math 12 -

this is a quadratic, it might be easier to see if you let
sinx = t, then we have

2t^2 + t - 6 = 0 which factors to
(2t - 3)(t + 2) = 0
t = 3/2 or t = -2

so sinx = 3/2 or sinx = -2

both cases are not possible since sinx must be between -1 and +1

so your equation has no solution.

• Math 12 -

let y = sin x
the equation then becomes:
2y^2 + y - 6 = 0
then factor:
(2y - 3)(y + 2) = 0
y = 3/2 and y = -2
therefore:
sin x = 3/2, x = 2*pi*n + arcsin 3/2 ; and
sin x = -2 , x = 2*pi*n + arcsin (-2) , where n = integer
*but note that arcsin 3/2 and arcsin (-2) are undefined.

so there,, :)

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