Wie löst man # cosx = 1 / 2 #?

#"since "cosx>0#

#"then x will be in the first/fourth quadrants"#

#cosx=1/2#

#rArrx=cos^-1(1/2)=pi/3larrcolor(blue)" angle in first quadrant"#

#"or "x=(2pi-pi/3)=(5pi)/3larrcolor(blue)" angle in fourth quadrant"#