Question

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

Answer 1

`x=(7pi)/12,(11pi)/12,(19pi)/12,(31pi)/12`

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`