# sin θ = -1

How to find the general solution of an equation of the form sin θ = -1?

Prove that the general solution of sin θ = -1 is given by θ = (4n - 1)π/2, n ∈ Z.

Solution:

We have,

sin θ = -1

⇒ sin θ = sin (-π/2)

θ = mπ + (-1)^m ∙ (-π/2), m ∈ Z, [Since, the general solution of sin θ = sin ∝ is given by θ = nπ + (-1)^n ∝, n ∈ Z.]

θ = mπ + (-1)^m ∙ π/2

Now, if m is an even integer i.e., m = 2n (where n ∈ Z) then,

θ = 2nπ - π/2

⇒ θ = (4n - 1) π/2 …………………….(i)

Again, if m is an odd integer i.e. m = 2n + 1 (where n ∈ Z) then,

θ = (2n + 1) ∙ π + π/2

⇒ θ = (4n + 3) π/2 …………………….(ii)

Now combining the solutions (i) and (ii) we get, θ = (4n - 1)π/2, n ∈ Z.

Hence, the general solution of sin θ = -1 is θ = (4n - 1)π/2, n ∈ Z.