# Factoring trig equations [Solved!]

**phinah** 15 Jul 2018, 16:44

### My question

In solving "`4\ sin^3x + 2\ sin^2 x - 2 sin x - 1 = 0`", can I factor as below?

### Relevant page

Analytic Trigonometry

### 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.

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.

## Re: Factoring trig equations

**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?

## Re: Factoring trig equations

**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.

## Re: Factoring trig equations

**Murray** 23 Jul 2018, 19:19

Looks good to me!

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