Solution of the equation sinx=5−14

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x=nπ+(−1)nπ10,n∈z

x=nπ+(−1)nπ5,n∈z

x=nπ+(−1)nπ12,n∈z

x=nπ+(−1)n2π5,n∈z

answer is A.

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Detailed Solution

sinx=5−14⇒sinx=sinπ10x=nπ+(−1)nπ10,n∈z

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