Question

How do you solve `sin(4x)=1`?

Answer 1

`x=(npi)/2+pi/8`, where `n` is an integer

Explanation:

As `sin4x=1=sin(pi/2)` and sine ratio repeats after every `2pi`

`4x=2npi+pi/2`, where `n` is an integer

or `x=(npi)/2+pi/8`, where `n` is an integer