No, 6sinx \(\displaystyle \not=\\) 0.
\(\displaystyle 6sinxcosx - cosx = 0\)
\(\displaystyle cosx(6sinx - 1) = 0\)
Either \(\displaystyle cosx = 0\) or \(\displaystyle 6sinx - 1 = 0\)
This equation seems to imply an infinite number of solutions, also.