We have,
√3cosθ+sinθ=1
Since, sinθ=√1−cos2θ
Therefore,
(1−√3cosθ)2=1−cos2θ
1+3cos2θ−2√3cosθ=1−cos2θ
4cos2θ−2√3cosθ=0
4cosθ=2√3
cosθ=2√34=√32
θ=cos−1√32=π6 or 2π−π6=11π6
But
θ=11π6 is only correct by putting in the equation.
Hence, this is the answer.