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Question: The general solution of \(2 \sin ^ { 2 } \theta - 3 \sin \theta - 2 = 0\)is...

The general solution of 2sin⁡2θ−3sin⁡θ−2=02 \sin ^ { 2 } \theta - 3 \sin \theta - 2 = 0is

A

nπ+(−1)nπ2n \pi + ( - 1 ) ^ { n } \frac { \pi } { 2 }

B

nπ+(−1)nπ6n \pi + ( - 1 ) ^ { n } \frac { \pi } { 6 }

C

nπ+(−1)n7π6n \pi + ( - 1 ) ^ { n } \frac { 7 \pi } { 6 }

D

nπ−(−1)nπ6n \pi - ( - 1 ) ^ { n } \frac { \pi } { 6 }

Answer

nπ−(−1)nπ6n \pi - ( - 1 ) ^ { n } \frac { \pi } { 6 }

Explanation

Solution

2sin⁡2θ−3sin⁡θ−2=02 \sin ^ { 2 } \theta - 3 \sin \theta - 2 = 0 ⇒2sin⁡2θ−4sin⁡θ+sin⁡θ−2=0⇒2sin⁡θ(sin⁡θ−2)+(sin⁡θ−2)=0\Rightarrow 2 \sin ^ { 2 } \theta - 4 \sin \theta + \sin \theta - 2 = 0 \Rightarrow 2 \sin \theta ( \sin \theta - 2 ) + ( \sin \theta - 2 ) = 0 (2sin⁡θ+1)(sin⁡θ−2)=0( 2 \sin \theta + 1 ) ( \sin \theta - 2 ) = 0

sin⁡θ=+2\sin \theta = + 2 (which is impossible) ⇒ ∴sin⁡θ=−12\therefore \sin \theta = - \frac { 1 } { 2 }

⇒ sin⁡θ=sin⁡(−π/6)⇒θ=nπ−(−1)nπ/6\sin \theta = \sin ( - \pi / 6 ) \Rightarrow \theta = n \pi - ( - 1 ) ^ { n } \pi / 6.