1. ## Trigonometry Help!

Question:
Find the solution of $\sin{x}-1=0$ if $0 \leq x < \frac{1}{2}\pi$.

Attempt:

$\sin{x}-1=0$

$\sin{x} = 1$

$x = \sin^{-1}{(1)}$

$x = 90^o$

$x = \frac{1}{2}\pi$

Either my answer is wrong or the question is wrong because it was mentioned in the question that it is not greater than $\frac{1}{2}\pi$. Am I correct?

2. You're correct, so the answer is that no real number solves this equation on the given interval.