Answer:
Given that,
[tex]1-\sin\theta=0\text{ for an interval of }\theta\text{ is 0}\leq\theta\leq2\pi[/tex]
On solving this we get,
[tex]\sin\theta=1[/tex]
we get,
[tex]\theta=\frac{\pi}{2}[/tex]
Since the period of sin theta is 2 pi.
The solution of the given equation is,
[tex]\theta=\frac{\pi}{2}[/tex]
