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

1. Apr 15, 2012

### okh

1. The problem statement, all variables and given/known data
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 [Broken]

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

2. Relevant equations

3. The attempt at a solution
For the first one:
http://img855.imageshack.us/img855/4633/imagekhd.jpg [Broken]

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

And I'm stuck with these cos^3x and sin^3x. I think I should replace something...
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited by a moderator: May 5, 2017
2. Apr 15, 2012

### SammyS

Staff Emeritus
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).

Last edited by a moderator: May 5, 2017
3. Apr 15, 2012

### okh

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.

4. Apr 15, 2012

### HallsofIvy

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.