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Questions  

General solution of 31cosθ+3+1sinθ=2 is

a
nπ+−1nπ4−π12
b
nπ2+π4
c
nπ+−1nπ4+π12
d
nπ+π4

detailed solution

Correct option is A

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

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