Why m? Your question equavalent to $sin3x=1,cos(60-x)=-1$ or $sin3x=-1,cos(60-x)=1$. It have not solution, because $sin3x=0$, when $cos(60-x)=\pm 1$.

The equation in the statement is missing a number '4'. The correct version should be $\frac{\sin 3x \cos (60^\circ -4x)+1}{\sin(60^\circ - 7x) - \cos(30^\circ + x) + m}=0$.

Note, that $\sin(3x) \cdot \cos(60-x)=-1,$ we have either, $\sin(3x)=-1, \cos(60-x)=-1,$ note that $\sin(270),$ is $-1,$ but $x=90,$ doesn't imply $\cos(-30)=-1.$ Now, if $\sin(3x)=1, \cos(60-x),$ then $x=30,$ but again, $\cos(30) \ne 1,$ so there are no solutions.

Find $x$ such that trigonometric \[\frac{\sin 3x \cos (60^\circ -4x)+1}{\sin(60^\circ - 7x) - \cos(30^\circ + x) + m}=0\]where $m$ is a fixed real number.