General solution of 3−1cosθ+3+1sinθ=2 is
nπ+−1nπ4−π12
nπ2+π4
nπ+−1nπ4+π12
nπ+π4
We have 3−1cosθ+3+1sinθ=2 ⇒sinθ3+122+cosθ3−122=222 ⇒sinθcosπ12+cosθsinπ12=12
⇒sinθ+π12=sinπ4 ⇒θ+π12=nπ+−1nπ4, n∈Z ⇒θ=nπ+−1nπ4-π12, n∈Z