phinah 15 Jul 2018, 16:44
In solving "`4\ sin^3x + 2\ sin^2 x - 2 sin x - 1 = 0`", can I factor as below?
`sin x(4\ sin^2x + 2 sin x - 2) - 1 = 0.`
I want to know if this is a correct way to factor the equation above.
X
In solving "`4\ sin^3x + 2\ sin^2 x - 2 sin x - 1 = 0`", can I factor as below?
Relevant page <a href="https://www.intmath.com/analytic-trigonometry/analytic-trigo-intro.php">Analytic Trigonometry</a> What I've done so far `sin x(4\ sin^2x + 2 sin x - 2) - 1 = 0.` I want to know if this is a correct way to factor the equation above.
Murray 16 Jul 2018, 01:38
@Phinah: While that is a correct factoring, it doesn't help you to solve it, whih I presume is what you need to do.
HINT: See the methods in this chapter: Polynomial Equations
It would be good to simplify things a bit and let, say `t = sin x`, and so the polynomial becomes:
`4 t^3 + 2 t^2 - 2t - 1`
As first guesses, `t=1` and `t=-1` don't work to make it have value `0.` Can you suggest another value?
X
@Phinah: While that is a correct factoring, it doesn't help you to solve it, whih I presume is what you need to do. HINT: See the methods in this chapter: <a href="https://www.intmath.com/equations-of-higher-degree/polynomial-equations.php">Polynomial Equations</a> It would be good to simplify things a bit and let, say `t = sin x`, and so the polynomial becomes: `4 t^3 + 2 t^2 - 2t - 1` As first guesses, `t=1` and `t=-1` don't work to make it have value `0.` Can you suggest another value?
phinah 23 Jul 2018, 11:46
Looking at it with `t = sin x` then it seems better to factor by grouping.
`4t^3`+2t^2-2t-1=0`
`2t^2(2t+1)-(2t+1)=0`
`(2t^2-1)(2t+1)=0`
`t= +- (sqrt 2)/2`
qq t = -1/2 qq
Thanks.
X
Looking at it with `t = sin x` then it seems better to factor by grouping. `4t^3`+2t^2-2t-1=0` `2t^2(2t+1)-(2t+1)=0` `(2t^2-1)(2t+1)=0` `t= +- (sqrt 2)/2` qq t = -1/2 qq Thanks.
X
Looks good to me!
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