Correct Answer - `theta=2n pi+(5 pi)/3, n in Z`
`(1-2 cos theta)^(2)+(tan theta +sqrt(3))^(2)=0`
`rArr (1-2 cos theta)^(2)=0`
and `(tan theta +sqrt(3))^(2)=0`
`rArr cos theta=1/2` and `tan theta=-sqrt(3)`
Hence, `theta` is in the fourth quadrant, and
`theta=2pi-pi/3=(5pi)/3`
Hence, the general solution is
`theta=2 npi+(5pi)/3, n in Z`.