Trig help - losing solutions when dividing

Hi,
When you have something like 6sinx/tanx=1
Ig you rearrange like this:
6sinx=tanx
6sinx=sinx/cosx
But I swear you can't divide in trig equations as dividing by sinx would miss the solutions for sinx=0 ?
So in this case why don't you have to do that?
Thank you

Reply 1

mqb2766

sin(x)=tan(x)=0 does not satisfy the original equation, so you can assume sin and tan are non-zero in the further analysis. In fact, its good practice to clearly state this.

You multiply throgh by tan() on line 2. You (probably) assume its non-zero at that point. The sin(x)=0 extraneous solution is introduced at this point.
https://en.wikipedia.org/wiki/Extraneous_and_missing_solutions#:~:text=In%20mathematics%2C%20an%20extraneous%20solution,valid%20solution%20to%20the%20problem.

The "obvious" way here is to use the identity for tan at the start and note its simply cos=1/6.

(edited 3 years ago)

Reply 2

Rhys_M

Original post by mqb2766

sin(x)=tan(x)=0 does not satisfy the original equation, so you can assume sin and tan are non-zero in the further analysis. In fact, its good practice to clearly state this.

You multiply throgh by tan() on line 2. You assume its non-zero at that point. The sin(x)=0 extraneous solution is introduced at this point.
https://en.wikipedia.org/wiki/Extraneous_and_missing_solutions#:~:text=In%20mathematics%2C%20an%20extraneous%20solution,valid%20solution%20to%20the%20problem.

The "obvious" way here is to use the identity for tan at the start and note its simply cos=1/6.

Oh so there are solutions for sinx=0 but at a different point?
How come, as this doesn't satisfy the original equation as 6sin0/tan0 is undefined

(edited 3 years ago)

Reply 3

mqb2766

The "solutions" sin(x)=0 are extraneous, so introduced as part of the working when you multiply through by tan(). They are not part of the original problem and you should explicitly reject them. The wiki page is decent. Their example is roughly solve
x+1 = 0
now multiply through by x
x^2 + x = 0
x(x+1) = 0
so "solutions" 0 and -1. Obviously when you multiply through by x, one extraneous solution will generally be x=0 as you'd have 0=0 after the mutliplicatoin, even when its not a solution to the original equation.

When you mulitply or divide an equation by an algebraic expression (or square or ...), the set of solutions of the before and after equations may be different.