Solving trigomonetry equation for x

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The discussion revolves around solving the trigonometric equation sin^3 x + cos^3 x + (1/4)(sin x - cos x) = cos 2x / (cos x - sin x). Participants initially derive the equation sin 4x - sin 2x + 1 = 0 but struggle with the next steps. A suggestion is made to substitute u = sin 2x to simplify the equation, although confusion arises regarding the equivalence of the original and derived equations. Ultimately, it is acknowledged that the expressions may not be equivalent, but their solution sets are similar, leading to the conclusion that finding a simple solution may not be feasible. The conversation highlights the complexity of solving such trigonometric equations without numerical methods.
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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
 
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Set u = sin2x. What's your new equation?
 
gb7nash said:
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...:confused:

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
 
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.
 
gb7nash said:
Starting from sin4x - sin2x + 1 = 0

How can you get that equation?
 
Edit:

My mistake, I thought you meant sin4x, not sin (4x). Ignore my last post!
 
gb7nash said:
Edit:

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

So, do you have new idea? :smile:

Or maybe it is not solvable...
 
You've made a mistake somewhere because the original equation and your final equation aren't equivalent.
 
Mentallic said:
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?
 
  • #10
Sorry, that's my mistake... The expressions aren't equivalent but the solution sets are :biggrin:
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
Mentallic said:
Sorry, that's my mistake... The expressions aren't equivalent but the solution sets are :biggrin:
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.
 

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