Question

$cos(0.5 pi -2x)+sinx=0$

Answer 1

$cos(0,5 pi -2x)+sinx=0 \ cos0,5 pi cos2x-sin0,5 pi sin2x+sinx=0 \ cos frac{ pi }{2} cos2x-sin frac{ pi }{2} sin2x+sinx=0 \ 0*cos2x-1*sin2x+sinx=0 \ sin2x+sinx=0 \ sinxcosx+sinx=0 \ sinx(cosx+1)=0 \ sinx=0 \ x= pi n \ cosx+1=0 \ cosx=-1 \ x= pi +2 pi n$