# How do I solve #sin 2x=-1/2# for #0<=x<=2pi#?

##### 1 Answer

Dec 2, 2017

#### Explanation:

#sin2x=-1/2#

#"since "sin2x<0" then x in third/fourth quadrants"#

#rArr2x=pi/6larrcolor(blue)"related acute angle"#

#"note "rarr0<=2x<=4pi#

#2x=(7pi)/6,(11pi)/6,(19pi)/6,(31pi)/6#

#rArrx=(7pi)/12" or "(11pi)/12" or "(19pi)/12" or "(31pi)/12#