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Question: The general solution of the equation \(( \sqrt { 3 } - 1 ) \sin \theta + ( \sqrt { 3 } + 1 ) \cos ...

The general solution of the equation

(3−1)sin⁡θ+(3+1)cos⁡θ=2( \sqrt { 3 } - 1 ) \sin \theta + ( \sqrt { 3 } + 1 ) \cos \theta = 2 is

A

2nπ±π4+π122 n \pi \pm \frac { \pi } { 4 } + \frac { \pi } { 12 }

B

nπ+(−1)nπ4+π12n \pi + ( - 1 ) ^ { n } \frac { \pi } { 4 } + \frac { \pi } { 12 }

C

2nπ±π4−π122 n \pi \pm \frac { \pi } { 4 } - \frac { \pi } { 12 }

D

nπ+(−1)nπ4−π12n \pi + ( - 1 ) ^ { n } \frac { \pi } { 4 } - \frac { \pi } { 12 }

Answer

2nπ±π4+π122 n \pi \pm \frac { \pi } { 4 } + \frac { \pi } { 12 }

Explanation

Solution

(3−1)sin⁡θ+(3+1)cos⁡θ=2( \sqrt { 3 } - 1 ) \sin \theta + ( \sqrt { 3 } + 1 ) \cos \theta = 2 Divided by

(3−1)2+(3+1)2=22\sqrt { ( \sqrt { 3 } - 1 ) ^ { 2 } + ( \sqrt { 3 } + 1 ) ^ { 2 } } = 2 \sqrt { 2 } in both sides,

We get, (3−1)22sin⁡θ+(3+1)22cos⁡θ=222\frac { ( \sqrt { 3 } - 1 ) } { 2 \sqrt { 2 } } \sin \theta + \frac { ( \sqrt { 3 } + 1 ) } { 2 \sqrt { 2 } } \cos \theta = \frac { 2 } { 2 \sqrt { 2 } }

sin⁡θsin⁡15∘+cos⁡θcos⁡15∘=12\sin \theta \sin 15 ^ { \circ } + \cos \theta \cos 15 ^ { \circ } = \frac { 1 } { \sqrt { 2 } }⇒ sin⁡θ⋅sin⁡π12+cos⁡θ⋅cos⁡π12=cos⁡π4\sin \theta \cdot \sin \frac { \pi } { 12 } + \cos \theta \cdot \cos \frac { \pi } { 12 } = \cos \frac { \pi } { 4 } cos⁡(θ−π12)=cos⁡π4\cos \left( \theta - \frac { \pi } { 12 } \right) = \cos \frac { \pi } { 4 }

⇒ θ−π12=2nπ±π4\theta - \frac { \pi } { 12 } = 2 n \pi \pm \frac { \pi } { 4 } ⇒ θ=2nπ±π4+π12\theta = 2 n \pi \pm \frac { \pi } { 4 } + \frac { \pi } { 12 }.