Come risolvi #cos 2x + sin x = 0 #?

#x=sin^-1(-1/2), x=sin^-1(1)#

Soluzione

#cos2x+sinx=0#

As

#cos2x=cos^2x-sin^2x#

So

#cos^2x-sin^2x+sinx=0#

#1-sin^2x-sin^2x+sinx=0#

#1-2sin^2x+sinx=0#

#-2sin^2x+sinx+1=0#

#2sin^2x-sinx-1=0#

#2sin^2x-2sinx+sinx-1=0#

#2sinx(sinx-1)+1(sinx-1)#

#(2sinx+1)(sinx-1)=0#

#(2sinx+1)=0, (sinx-1)=0#

#2sinx=-1, sinx=1#

#sinx=-1/2, sinx=1#

#x=sin^-1(-1/2), x=sin^-1(1)#