Question

Please help me solve the equation on the interval shown in the picture. Must show work.

Answer 1

Given that,

`1-\sin\theta=0\text{ for an interval of }\theta\text{ is 0}\leq\theta\leq2\pi`

On solving this we get,

`\sin\theta=1`

we get,

`\theta=(\pi)/(2)`

Since the period of sin theta is 2 pi.

The solution of the given equation is,

`\theta=(\pi)/(2)`