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Factoring: cos^3t + cos2t + sin^3 t = 0

Date: 04/23/2003 at 17:29:35
From: Tony
Subject: Trigonometric Equations

cos^3t + cos2t + sin^3 t = 0

How can this expression equal 0?  I need to find ALL solutions between 
0 and pi (unit circle).

Date: 04/25/2003 at 09:51:51
From: Doctor Tracy
Subject: Re: Trigonometric Equations

Tony, thanks for writing in.  

One of the hardest things to learn about doing mathematics is that you 
don't always know how the problem is going to turn out when you start 
it. When you're solving an equation, as you're doing here, there are 
many things that you're *allowed* to do to solve it. For example, you 
can add 1 to both sides of the equation. In this case, adding 1 to 
both sides of the equation probably won't be helpful in solving it, 
but that doesn't make it wrong. You just have to try whatever you can 
think of.  

For the problem that you wrote in about, I tried a lot of things that 
didn't help, but in the end I came up with the answer. I'll talk you 
through what I did (leaving out the extraneous/unhelpful steps):

The original equation is 

      3                3
   cos t + cos 2t + sin t = 0  

As a general strategy, I figure that I probably want to keep the 0 on 
the right-hand side and just want to try to factor the left-hand side.  
If I can completely factor the left-hand side, then I can solve the 
equation by determining the values of t that make each factor zero. 
This is basically the method you use for solving polynomial equations.  
For example, if you want to solve x^2 + 3x + 2 = 0, you factor the 
left-hand side: (x + 1)(x + 2) = 0 and determine what values make each 
factor zero: x = -1 and x = -2.

The first thing I noticed with this equation was that I can use the 
trig identity 

              2       2 
  cos 2t = cos t - sin t 

to rewrite this as:
     3       2       2       3
  cos t + cos t - sin t + sin t = 0.

Now there are a lot of things that you can try at this point. I'm 
trying to factor this, so I tried various groupings of these terms to 
see if anything worked. The grouping that worked for me was this:

      3       3         2       2
  (cos t + sin t) + (cos t - sin t) = 0.

Now I'll use the 'sum of cubes' and 'difference of squares' 
factorization formulas I learned in algebra to factor these. Remember
that a sum of cubes factors as

   3   3           2        2
  a + b = (a + b)(a - ab + b ),

and a difference of squares factors as

   2   2
  a - b = (a + b)(a - b)

Using these we can factor each grouping above to get
                           2                        2                 
  0 =   (cos t + sin t)(cos t - (cos t)(sin t) + sin t) 
      + (cos t + sin t)(cos t - sin t) 

Using the fact that

     2       2
  cos t + sin t = 1, 

I can rewrite the equation as:
  0 =   (cos t + sin t)(1 - (cos t)(sin t)) 
      + (cos t + sin t)(cos t - sin t)

Now, notice that there is a common factor of (cos t + sin t). 
Factoring that out gives:

  0 = (cos t + sin t)(1 - (cos t)(sin t) + cos t - sin t) 

At this point we've made some good progress in our plan to factor the 
left-hand side of the equation. The first factor, cos t + sin t, is 
completely factored, but we still have to do some work on the second 

We can factor this using the technique of factoring by grouping, but
we'll need to rearrange the terms first. Let's rewrite our equation as

  0 = (cos t + sin t)((1 - sin t) + (cos t - (cos t)(sin t))) 


  0 = (cos t + sin t)((1 - sin t) + (cos t)(1 - sin t)) 

Factoring out the common factor of (1 - sin t) gives

  0 = (cos t + sin t)(1 - sin t)(1 + cos t) 

Now we have completed the factoring.  So,

  cos t + sin t = 0  or  1 - sin t = 0  or  1 + cos t = 0.

These are relatively easy equations to solve, so I'll let you finish 
from here yourself.

I hope this helps, and I hope you see that, at each step of the 
problem, I was just looking for anything that I could use to rewrite 
the equation. I never looked too far ahead other than keeping in mind 
the idea that I wanted to factor the left-hand side of the equation. 

Please write in if I can be of any further help.

- Doctor Tracy, The Math Forum 
Associated Topics:
High School Trigonometry

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