Math 12
posted by H .
trig question
2sin^2x+sinx6=0
how to solve for x when the
restriction is x is greater than or equal to zero but less than 2pi

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. 
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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