Find the general solution of the following equation cot2 β+ [3/sin β] +3=0? Fast please and accurate answer only.
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Correct option isAnπ+(−1) n (− 6π )Given cot 2 θ+ sinθ3 +3=0As we know that cot 2 θ=cosec 2 θ−1cosec 2 θ−1+3cosecθ+3=0⟹cosec 2 θ+3cosecθ+2=0⟹(cosecθ+2)(cosecθ+1)=0⟹cosecθ=−2 or −1So x=nπ+(−1) n
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