How do you solve #sin x=-sqrt2/2#?
1 Answer
Mar 6, 2018
Explanation:
#"since "sinx<0" then x is in third/fourth quadrant"#
#rArrx=sin^-1(sqrt2/2)=pi/4larrcolor(red)"related acute angle"#
#rArrx=pi+pi/4=(5pi)/4larrcolor(red)"in third quadrant"#
#rArrx=2pi-pi/4=(7pi)/4larrcolor(red)"in fourth quadrant"#
#"since sin is periodic these solutions will be repeated"#
#"every"2pi#
#rArrx=(5pi)/4+2kpitok inZZ#
#rArrx=(7pi)/4+2kpitok inZZ#
