# Solving trigomonetry equation for x

1. Dec 4, 2011

### songoku

1. The problem statement, all variables and given/known data
Solve
$$sin^3 x + cos^3 x + \frac{1}{4}(sin x - cos x) = \frac{cos 2x}{cos x - sin x}$$

2. Relevant equations
trigonometry

3. The attempt at a solution
After putting some effort, I got: sin 4x - sin 2x + 1 = 0

I don't know how to proceed....

Thanks

2. Dec 4, 2011

### gb7nash

Set u = sin2x. What's your new equation?

3. Dec 4, 2011

### songoku

I don't get your hint.

sin 4x - sin 2x + 1 = 0
2 sin 2x cos 2x - 2 sin x cos x + 1 = 0
4 sin x cos x (1 - 2 sin2x) - 2 sin x cos x + 1 = 0
4 sin x cos x (1 - 2u) - 2 sin x cos x + 1 = 0

and then...

Should I change u = sin2x to sin x = √u then draw triangle to find cos x in term of u? I think it will be more complicated

4. Dec 4, 2011

### gb7nash

You're thinking into this way too much.

Starting from sin4x - sin2x + 1 = 0, make a simple substitution u = sin2x and plug the u stuff into the equation.

5. Dec 4, 2011

### songoku

How can you get that equation?

6. Dec 4, 2011

### gb7nash

Edit:

My mistake, I thought you meant sin4x, not sin (4x). Ignore my last post!

7. Dec 4, 2011

### songoku

So, do you have new idea?

Or maybe it is not solvable...

8. Dec 4, 2011

### Mentallic

You've made a mistake somewhere because the original equation and your final equation aren't equivalent.

9. Dec 4, 2011

### songoku

$$sin^3 x + cos^3 x + \frac{1}{4}(sin x - cos x) = \frac{cos 2x}{cos x - sin x}$$

$$(sin x + cos x) (sin^2 x - sin x cos x + cos^2 x) - \frac{1}{4}(cos x - sin x) = \frac{cos 2x}{cos x - sin x}$$

$$cos 2x (1 - sin x cos x) - \frac{1}{4}(1 - sin 2x) = cos 2x$$

$$4 cos 2x sin x cos x + 1 - sin 2x = 0$$

$$sin 4x - sin 2x + 1 = 0$$

Correct?

10. Dec 4, 2011

### Mentallic

Sorry, that's my mistake... The expressions aren't equivalent but the solution sets are
minus the $x\neq n\pi+\pi/4$ of course.

It doesn't seem as though this is simple to solve. Were you expecting a numerical solution?

11. Dec 4, 2011

### songoku

I don't think so; we haven't covered numerical solution.