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Question

General solutions of x for which 2sinx+1=0 is :

A
nπ+(1)n(π6), nZ
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B
nπ±(1)n2π3, nZ
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C
nπ±π3, nZ
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D
nπ±(1)n(π6), nZ
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Solution

The correct option is A nπ+(1)n(π6), nZ
2sinx+1=0sinx=12sinx=sin(π6)x=nπ+(1)n(π6), nZ

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