How do you solve #cosx=1/2#?
1 Answer
Sep 19, 2017
Explanation:
#"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"#
