Find the general solutions from the following equations:
i) `sintheta=-1/2`
ii) `cos=-sqrt(3)/2`
`tantheta=-sqrt(3)`

Explanation

i) `sintheta=1-1/2=-sinpi/6=sin(-pi/6)`
`therefore =npi+(-1)^n.(-pi/6)`
`rArr theta=npi-(-1)^(n)pi/6`. Ans
ii) `costheta=-sqrt(3)/2=-cospi/6`
`=cos(pi-pi/6)=cos(5pi)/(6)`
`therefore theta=2npi+-(5pi)/(6)` Ans.
iii) `tantheta=-sqrt(3)`
`=-tanpi/3=tan(-pi/3)`
`therefore theta=npi-pi/3`. Ans.