Question

How do you solve `sin x=-sqrt2/2`?

Answer 1

`"see explanation"`

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`