Question:

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

Attempt:

$\displaystyle \sin{x}-1=0$

$\displaystyle \sin{x} = 1$

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

$\displaystyle x = 90^o$

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

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