Solving trigomonetry equation for x

[songoku]

Homework Statement

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

Homework Equations

trigonometry

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

 

[gb7nash]

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

 

[songoku]

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

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

 

[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.

 

[songoku]

Starting from sin⁴x - sin²x + 1 = 0

How can you get that equation?

 

[gb7nash]

Edit:

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

 

[songoku]

Edit:

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

So, do you have new idea?

Or maybe it is not solvable...

 

[Mentallic]

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

 

[songoku]

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

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

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

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

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

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

Correct?

 

[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?

 

[songoku]

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?

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