C2 Trig help?

Why can't you divide both sides by tan x to get:
Cos x =1

Sin x = tan x

Why can't you divide both sides by tan x to get:
Cos x =1
For which the solutions are 0 , 2pi etc

Because u lose a solution. You should always look to factorise when solving trig equations. for this question:
- convert tanx into the identity sinx/cosx
-sinx=sinx/cosx
-multiply both sides by cosx to get rid of denominator
-sinxcosx=sinx
-take away sinx from both sides
-sinxcosx-sinx=0
-now take out the common factor of sinx
-sinx (cosx-1)=0
-so either sinx=0, or cosx-1=0

As you mentioned, you get one of the solutions as being cosx=1 (cosx-1=0, cosx=1). However theres a second solution of sinx=0 which you would have missed had you not factorised. So just remember u should always look to factorise

Because tan x might be equal to zero, and you always need to be sure that you are not dividing by zero.

Sin x = tan x

Why can't you divide both sides by tan x to get:
Cos x =1
For which the solutions are 0 , 2pi etc

So you never divide by cos, tan or sin because there is a chance they could be zero?

This would be a good rule to follow, unless there is something specific in the question that makes it clear that they will never be zero in this particular question.

It's not that you can't. The problem is that doing so could cause you to lose solutions. $\sin x = \tan x \Rightarrow \sin x = \frac{\sin x}{\cos x} \Rightarrow \frac{\sin x \cos x}{\sin x}=1$. Now you have to consider the cases where $\sin x = 0$ which will occur every $m\pi|m\in\mathbb{Z}$. So, you could potentially be dividing by 0.

I can't understand that part. How can a question tell you whether tan cos sin is not zero?

This could happen if you are only told to consider particular values of x. For example, if your question had been to solve sin x = tan x for 0 < x < 180 (note the strict inequalities), you could divide by tan x becasue it is never zero for these values of x.

Because u lose a solution. You should always look to factorise when solving trig equations. for this question:
- convert tanx into the identity sinx/cosx
-sinx=sinx/cosx
-multiply both sides by cosx to get rid of denominator
-sinxcosx=sinx
-take away sinx from both sides
-sinxcosx-sinx=0
-now take out the common factor of sinx
-sinx (cosx-1)=0
-so either sinx=0, or cosx-1=0

As you mentioned, you get one of the solutions as being cosx=1 (cosx-1=0, cosx=1). However theres a second solution of sinx=0 which you would have missed had you not factorised. So just remember u should always look to factorise

Because u lose a solution. You should always look to factorise when solving trig equations. for this question:
- convert tanx into the identity sinx/cosx
-sinx=sinx/cosx
-multiply both sides by cosx to get rid of denominator
-sinxcosx=sinx
-take away sinx from both sides
-sinxcosx-sinx=0
-now take out the common factor of sinx
-sinx (cosx-1)=0
-so either sinx=0, or cosx-1=0

As you mentioned, you get one of the solutions as being cosx=1 (cosx-1=0, cosx=1). However theres a second solution of sinx=0 which you would have missed had you not factorised. So just remember u should always look to factorise

So if there is no possible way to factorise e.g. 3sin x = cos x, You can cancel using identies right?