Htykuut

Assuming this is correct:

$arcsin x neq 0$

$sin(0) neq x$

$0neq x$

Following the same logic, why is this incorrect?

$arcsin(x+frac{1}{3}) geq 0$

$sin (0) geq x+frac{1}{3}$

$0 geq x + frac{1}{3}$

$x leq -frac{1}{3}$

$xin(-infty; -frac{1}{3}] $

The correct answer should be:

$xin[-frac{1}{3}; infty)$

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 $

$ expressiongeq 0 $

and

$arccos (expression) geq 0 $

$ expressiongeq 0 $

Thanks

The arcsine is an increasing function, so $arcsin agearcsin b$ is equivalent to $age b$ (provided $a$ and $b$ are in the domain of the arcsine).

Thus your inequality $arcsin(x+1/3)ge0=arcsin 0$ becomes $x+1/3ge0$, that is, $xge-1/3$. Taking into account that $-1le x+1/3le 1$, we get $-4/3le xle 2/3$, so combining the inequalities yields
$$
-frac{1}{3}le xlefrac{2}{3}
$$
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 $arccos agearccos b$ becomes $ale b$. Since $0=arccos(1)$, the inequality $arccos xge 0$ becomes $xle1$. This is anyway obvious, because $arccos xge0$ for every $xin[-1;1]$.