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Question: If \(\sqrt { 3 } \cos \theta + \sin \theta = \sqrt { 2 }\) , then general value of \(\theta\) is...

If 3cos⁡θ+sin⁡θ=2\sqrt { 3 } \cos \theta + \sin \theta = \sqrt { 2 } , then general value of θ\theta is

A

nπ+(−1)nπ4n \pi + ( - 1 ) ^ { n } \frac { \pi } { 4 }

B

(−1)nπ4−π3( - 1 ) ^ { n } \frac { \pi } { 4 } - \frac { \pi } { 3 }

C

nπ+π4−π3n \pi + \frac { \pi } { 4 } - \frac { \pi } { 3 }

D

nπ+(−1)nπ4−π3n \pi + ( - 1 ) ^ { n } \frac { \pi } { 4 } - \frac { \pi } { 3 }

Answer

nπ+(−1)nπ4−π3n \pi + ( - 1 ) ^ { n } \frac { \pi } { 4 } - \frac { \pi } { 3 }

Explanation

Solution

3cos⁡θ+sin⁡θ=2\sqrt { 3 } \cos \theta + \sin \theta = \sqrt { 2 } ⇒32cos⁡θ+12sin⁡θ=12\Rightarrow \frac { \sqrt { 3 } } { 2 } \cos \theta + \frac { 1 } { 2 } \sin \theta = \frac { 1 } { \sqrt { 2 } }

⇒sin⁡π3cos⁡θ+cos⁡π3sin⁡θ=12\Rightarrow \sin \frac { \pi } { 3 } \cos \theta + \cos \frac { \pi } { 3 } \sin \theta = \frac { 1 } { \sqrt { 2 } } ⇒sin⁡(θ+π3)=sin⁡π4\Rightarrow \sin \left( \theta + \frac { \pi } { 3 } \right) = \sin \frac { \pi } { 4 } ⇒θ=nπ+(−1)nπ4−π3\Rightarrow \theta = n \pi + ( - 1 ) ^ { n } \frac { \pi } { 4 } - \frac { \pi } { 3 }.