What are the solutions of sin (x – π) + 2 = 1 in [0, 2π)?

Question

Trigonometry question

Answer 1

sin(x-π)=sinxcosπ-coxsinπ=-sinx

siπ(x - π) + 2 = 1 now reads -sinx+2=1 or sinx=1 so

in [0,2π] the only answer is x=π/2

Answer 2

If sin(x - π) + 2 = 1, then sin(x - π) = -1

 

So, x - π = 3π/2 + 2kπ, where k = 0, ±1, ±2, ...

 

          x = 5π/2 + 2kπ

 

If k = -1, then x = π/2

 

Other choices for k give values of x that lie outside of the interval [0, 2π).