Q.
Solution of the equation sinx=5−14
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a
x=nπ+(−1)nπ10,n∈z
b
x=nπ+(−1)nπ5,n∈z
c
x=nπ+(−1)nπ12,n∈z
d
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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