The set of solutions of the equation (3−1)sin⁡θ+(3+1)cos⁡θ=2 is

(3−1)sin⁡θ+(3+1)cos⁡θ=2⇒(3−1)22sin⁡θ+(3+1)22cos⁡θ=222 ⇒cos θ cos π12+sin θ sinπ12=12 ⇒cos θ- π12=cos π4⇒θ=2nπ±π4+π12

The set of solutions of the equation (3−1)sin⁡θ+(3+1)cos⁡θ=2 is

(3−1)sin⁡θ+(3+1)cos⁡θ=2⇒(3−1)22sin⁡θ+(3+1)22cos⁡θ=222 ⇒cos θ cos π12+sin θ sinπ12=12 ⇒cos θ- π12=cos π4⇒θ=2nπ±π4+π12

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The set of solutions of the equation $(\sqrt{3} - 1) \sin θ + (\sqrt{3} + 1) \cos θ = 2$ is

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$\{2 nπ \pm \frac{π}{4} + \frac{π}{12} : n \in Z\}$

$\{2 nπ \pm \frac{π}{4} - \frac{π}{12} : n \in Z\}$

$\{nπ + (- 1)^{n} \frac{π}{4} + \frac{π}{12} : n \in Z\}$

$\{nπ + (- 1)^{n} \frac{π}{4} - \frac{π}{12} : n \in Z\}$

answer is A.

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Detailed Solution

$\begin{array}{l} (\sqrt{3} - 1) \sin θ + (\sqrt{3} + 1) \cos θ = 2 \\ \Rightarrow \frac{(\sqrt{3} - 1)}{2 \sqrt{2}} \sin θ + \frac{(\sqrt{3} + 1)}{2 \sqrt{2}} \cos θ = \frac{2}{2 \sqrt{2}} \Rightarrow \cos θ \cos \frac{π}{12} + \sin θ \sin \frac{π}{12} = \frac{1}{\sqrt{2}} \Rightarrow \cos (θ - \frac{π}{12}) = \cos \frac{π}{4} \\ \Rightarrow θ = 2 nπ \pm \frac{π}{4} + \frac{π}{12} \end{array}$