The general solution of sin = 0 is(a) nπ where n is a real number(b)

1.	The general solution of sin = 0 is(a) nπ where n is a real number(b) nπ, where n is an integer(c) 2π(d) π
Answer» $\implies$ we know that if sin X = 0, then x = 0, π, 2π, π, -2π, -6π, ETC. are SOLUTIONS of the given EQUATION. Hence, the general solution for sin x = 0 will be, x = nπ, where n∈I.

The general solution of sin = 0 is(a) nπ where n is a real number(b) nπ, where n is an integer(c) 2π(d) π

Answer»

$\implies$

we know that if sin X = 0, then x = 0, π, 2π, π, -2π, -6π, ETC.
are SOLUTIONS of the given EQUATION.
Hence, the general solution for sin x = 0 will be, x = nπ, where n∈I.

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The general solution of sin = 0 is(a) nπ where n is a real number(b) nπ, where n is an integer(c) 2π(d) π

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