Question

Find the solution of `2sqrt3sin2theta=sqrt3`, if `0<=theta<2pi`?

Answer 1

`theta=pi/12,(5pi)/12,(13pi)/12,(17pi)/12`

Explanation:

`2sqrt3sin2theta=sqrt3`

`rArrsin2theta=sqrt3/(2sqrt3)=1/2`

`sin2theta>0totheta" in first/second quadrants"`

`rArr2theta=pi/6" or "(5pi)/6" or "(13pi)/6" or "(17pi)/6`

`rArrtheta=pi/12" or "(5pi)/12" or "(13pi)/12" or "(17pi)/12to[0,2pi]`