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.
11 and 12 Grade Math
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