Solve this inequality: $\sin x<1$

Question

it may be a simple question, but as simple as it is I want to know how to solve it: $$\sin {x}<1$$ What is the solution of this inequality? Thank you!

Answer 1

We have $$\cos^2x=1-\sin^2x$$

For real $\displaystyle x,\cos^2x\ge0\implies 1-\sin^2x\ge0\iff -1\le \sin x\le1$

Conversely, if $\displaystyle\sin x>1, \sin^2x>1,\cos^2x=1-\sin^2x<0$

In fact, $\displaystyle\sin x=1\implies =2n\pi+\frac\pi2$ where $n$ is any integer

So, $\displaystyle x\ne2n\pi+\frac\pi2$

Answer 2

The solution to $$\sin x = 1$$ is $\left\{\ldots,\frac{-11\pi}{2},\frac{-7\pi}{2},\frac{-3\pi}{2},\frac{\pi}{2},\frac{5\pi}{2},\frac{9\pi}{2},\ldots\right\}$ (why?).

So the solution set to $$\sin x < 1$$ should be $$\mathbb{R}\setminus\left\{\ldots,\frac{-11\pi}{2},\frac{-7\pi}{2},\frac{-3\pi}{2},\frac{\pi}{2},\frac{5\pi}{2},\frac{9\pi}{2},\ldots\right\}$$

since, for real $x$, we have $\sin x < 1$ if just $\sin x\ne 1$.