# Factoring trig equations (2) [Solved!]

**phinah** 25 Jul 2018, 15:30

### My question

With an equation of the form 2 `sin^2` x + sin 2x = 0, I need help in determining a first step to solving it.

### Relevant page

5. Trigonometric Equations

### What I've done so far

Tried to go the route of (2 sin x + ....)(sin x + .... ) but saw immediately it is a no-go.

X

With an equation of the form 2 `sin^2` x + sin 2x = 0, I need help in determining a first step to solving it.

Relevant page
<a href="https://www.intmath.com/analytic-trigonometry/5-trigonometric-equations.php">5. Trigonometric Equations</a>
What I've done so far
Tried to go the route of (2 sin x + ....)(sin x + .... ) but saw immediately it is a no-go.

## Re: Factoring trig equations (2)

**Murray** 26 Jul 2018, 21:28

My hint is to remind yourself of the first formula on this page: Double Angle Formulas

X

My hint is to remind yourself of the first formula on this page: <a href="https://www.intmath.com/analytic-trigonometry/3-double-angle-formulas.php">Double Angle Formulas</a>

## Re: Factoring trig equations (2)

**phinah** 21 Aug 2018, 15:44

Okay thanks for that hint.

I substituted `2 sin x cos x` for `sin 2x.`

So `2\ sin^2 x + 2 sin x cos x = 0;`

Factoring: `2 sin x (sin x + cos x) = 0;

`

`sin x = 0` or `sin x + cos x = 0;`

For `sin x = 0:` `x = 0,` `pi,` `2pi`;

For `sin x + cos x = 0,` `sin x = - cos x:` `x = (3pi)/4.`

X

Okay thanks for that hint.
I substituted `2 sin x cos x` for `sin 2x.`
So `2\ sin^2 x + 2 sin x cos x = 0;`
Factoring: `2 sin x (sin x + cos x) = 0;
`
`sin x = 0` or `sin x + cos x = 0;`
For `sin x = 0:` `x = 0,` `pi,` `2pi`;
For `sin x + cos x = 0,` `sin x = - cos x:` `x = (3pi)/4.`

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