·       Ta có:sin4x−cos3x=sin4x−sinπ2−3x

=2cosx2+π4sin7x2−π4                                       

·       Điều kiện:cos5x≠0cosx2+π4≠0sin7x2+π4≠0⇔x≠π10+kπ5x≠π2+k2πx≠−π14+k2π7

·       Vậy TXĐ:D=ℝ\π10+kπ5;π2+k2π,−π14+k2π7 .