We have
`sin theta + sqrt(3) cos theta ge 1, -pi lt theta le pi`
`:. 1/2 sin theta + sqrt(3)/2 cos theta ge 1/2`
`rArr cos (theta-pi//6) ge 1/2`
`rArr cos x ge 1/2`, where `x= theta-pi//6`
`:. X in (- (7pi)/6, (5pi)/6)`
Draw the graph of `y = cos x` and `y=1/2` as shown in the following figure.
From the figure, `x in (- pi/3, pi/3)`
`:. theta in (-pi/6, pi/2)`