How to simplify trigonometric equations? (see details of question!)

Question

This is the questions I'm stuck on: 

 

Solve algebraically: f(x) = g(x) when f(x) = 4sin(2x)cos(2x)+2 and g(x)= 4sin(3x)sin(2x) + 2

 

1. I equated the equations

         4 sin(2x)cos(2x)+2 = 4sin(3x)sin(2x) + 2

 

2. I simplified the equation 

         sin(2x)cos(2x)+2 = sin(3x)sin(2x) + 2

          sin(2x)cos(2x)    = sin(3x)sin(2x) 

                    cos(2x)    = sin(3x)

 

3. Now I go on to solve cos(2x) = sin(3x)

    BUT the answers say there is a SECOND EQUATION to solve: sin(2x) = 0

 

    I thought I canceled sin(2x) out when I simplified the equation, so I wouldn't have to deal with it. 

    Please help me understand why this equation is important, and moreover, why it is equal to 0! 

 

    I can solve the rest of the question, I understand how to do that, but really need help understanding where the sin(2x) = 0        comes from! Thank you! 

 

 

         

Answer 1

I'll use a simpler example to explain what's happening.  Suppose you were trying to solve the equation x² = 3x.  If you divided both sides by x, you would get the solution x = 3.  But you would lose the solution x = 0 (which is obviously a valid solution for the original equation).  Why?  Because division by 0 is undefined, so you can only divide both sides of an equation by a non-zero value.  So anytime you divide both sides of an equation by a variable you must account for the case that the variable might be equal to 0.  In fact, it's usually a good idea to try to avoid dividing both sides of an equation by a variable so you don't have to worry about any special cases.  You can solve without dividing by bringing all the terms to the left side and factoring:  x² - 3x = 0, so x(x - 3) = 0, so x = 0 and x = 3 are the solutions.  

 

If you do the same for your problem you will get sin(2x)cos(2x) - sin(3x)sin(2x) = 0, which factors to 

 

[sin(2x)]*[cos(2x) - sin(3x)] = 0  

 

which implies that sin(2x) = 0 or cos(2x) = sin(3x).