Assuming this is correct:
$\arcsin x \neq 0$
$\sin(0) \neq x$
$0\neq x$
Following the same logic, why is this incorrect?
$\arcsin(x+\frac{1}{3}) \geq 0$
$\sin (0) \geq x+\frac{1}{3}$
$0 \geq x + \frac{1}{3}$
$x \leq -\frac{1}{3}$
$x\in(-\infty; -\frac{1}{3}] $
The correct answer should be:
$x\in[-\frac{1}{3}; \infty)$
Why is this incorrect? Can I use the $ \sin(0) $ trick only for when there's $\neq$ sign, thus cannot be used in inequalities?
I do have one more question, by the way. Is this correct?
$ arcsin(expression) \geq 0 $
$ expression\geq 0 $
and
$arccos (expression) \geq 0 $
$ expression\geq 0 $
Thanks
The arcsine is an increasing function, so $\arcsin a\ge\arcsin b$ is equivalent to $a\ge b$ (provided $a$ and $b$ are in the domain of the arcsine).
Thus your inequality $\arcsin(x+1/3)\ge0=\arcsin 0$ becomes $x+1/3\ge0$, that is, $x\ge-1/3$. Taking into account that $-1\le x+1/3\le 1$, we get $-4/3\le x\le 2/3$, so combining the inequalities yields $$ -\frac{1}{3}\le x\le\frac{2}{3} $$ There's no way the solution set is $[-1/3;\infty)$
For the arccosine you have to take into account that it is decreasing, so $\arccos a\ge\arccos b$ becomes $a\le b$. Since $0=\arccos(1)$, the inequality $\arccos x\ge 0$ becomes $x\le1$. This is anyway obvious, because $\arccos x\ge0$ for every $x\in[-1;1]$.