Trigonometric Equations - Why Do I Always Get cos^3 or sin^3?

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Homework Statement


I'm trying to solve two similar equations, but I can't go on.
This is the first one
http://img818.imageshack.us/img818/298/imageevsu.jpg

This is the second one:
http://img7.imageshack.us/img7/7812/imageowud.jpg

Homework Equations





The Attempt at a Solution


For the first one:
http://img855.imageshack.us/img855/4633/imagekhd.jpg

And for the second equation:
http://img12.imageshack.us/img12/2990/imageiag.jpg

And I'm stuck with these cos^3x and sin^3x. I think I should replace something...
 
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okh said:

Homework Statement


I'm trying to solve two similar equations, but I can't go on.
This is the first one
http://img818.imageshack.us/img818/298/imageevsu.jpg

This is the second one:
http://img7.imageshack.us/img7/7812/imageowud.jpg

Homework Equations



The Attempt at a Solution


For the first one:
http://img855.imageshack.us/img855/4633/imagekhd.jpg

And for the second equation:
http://img12.imageshack.us/img12/2990/imageiag.jpg

And I'm stuck with these cos^3x and sin^3x. I think I should replace something...
For the first one, change sin2(x) to 1-cos2(x). You will have a cubic equation in cos(x).

For the second one, change cos2(x) to 1-sin2(x). You will have a cubic equation in sin(x).
 
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Okay, but how to solve cubic trigonometic equations?
I think I can get a quadratic equation, in some way. We are studying them, indeed, and there is nothing about cubic trigonometric equations in my book.
 
Well, if you don't do something you can't possibly get an answer!

Have you tried just checking some simple values? It helps sometimes to know the "rational root theorem". Any rational root, of the form p/q with integers p and q, of the polynoial a_nx^n+ a_{n-1}x^{n-1}+ a_1x+ a_0= 0 must have p evenly dividing a_0 and q evenly dividing a_n. That will reduce the number of trials. Of course, it is not necessary that a polynomial equation have rational roots but this one does.

Once you have a single root, say x= a, divide the cubic polynomial by x- a to reduce to a quadratic.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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