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# Factoring trig equations (2) [Solved!]

### 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)

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)

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.