Solution of the equation sinx=5−14
x=nπ+(−1)nπ10,n∈z
x=nπ+(−1)nπ5,n∈z
x=nπ+(−1)nπ12,n∈z
x=nπ+(−1)n2π5,n∈z
sinx=5−14⇒sinx=sinπ10