Question

sen(6x-36)-cos(2x+46=0 resolver porfa

Answer 1

$\displaystyle \sin(6x-36)=\cos(2x+46)\\ \\ \sin(6x-36)=\sin[(2x+46)-90(2k+1)]\;;\forall k\in \mathbb Z\\ \\ 6x-36=(2x+46)-90(2k+1) \\ \\ 4x-82=-90(2k+1)\\ \\ 4x=-180k -8\\ \\ \boxed{x=-45k -2}$