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# Factoring trig equations [Solved!]

### 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

@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

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

Looks good to me!

X

Looks good to me!