A level maths trigonometry question

I was solving this trigonometric equation: -sin3x/cos3x =4sin3x and I multiplied cos3x on both sides to give
-Sin3x= 4sin3xcos3x. Then I went on to divide by 4sin3x on both sides however the markscheme said that it was wrong and that you had to subtract sin3x on both sides. Do you know why I can’t divide by sin3x to solve for x? Thanks
Original post by abibabymelon
I was solving this trigonometric equation: -sin3x/cos3x =4sin3x and I multiplied cos3x on both sides to give
-Sin3x= 4sin3xcos3x. Then I went on to divide by 4sin3x on both sides however the markscheme said that it was wrong and that you had to subtract sin3x on both sides. Do you know why I can’t divide by sin3x to solve for x? Thanks

Looking at the original equation, sin(3x) = 0 is obviously a solution as 0=0. So by dividing by sin(3x) you lose that solution. By taking it over and factorising, as per the mark scheme, you identify that factor.

Note you could divide by sin(3x) and simply note that it assumes sin(3x)!=0 and handle the case seperately. However, its easy to forget that you need to do this, so factorising rather than dividing is generally the best approach.