Posted by **H** on Thursday, October 21, 2010 at 12:55am.

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 -
**Reiny**, Thursday, October 21, 2010 at 8:43am
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 -
**jai**, Thursday, October 21, 2010 at 8:57am
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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