Find all the values for $x$: $\sin(x) < 0,6$

Question

$\sin(x) < 0.6$

My solution is

$x<\arcsin(0.6) +2kπ$ OR $x<\pi-\arcsin(0.6) +2k\pi$

$x<0.6435+2k\pi$ OR $x<2.4981+2k\pi$

This is correct but how can I put these in an interval? Also when I pick $\frac{\pi}2$ which is smaller than $2.4981$ the sinus is $1$???

Answer 1

Just use common sense.

$\sin x$ starts at $x=0; \sin x=0$ and it immediately increases as $x$ increases. At $x = \arcsin .6$ then $\sin x= .6$. $0 < \arcsin .6 < \frac \pi 2$. Let's call $\arcsin .6 = \alpha$. So that's our first interval: For $x \in [0, \alpha), \sin x < .6$.

And for all $k \in \mathbb Z$ we get $x \in [2k\pi, 2k\pi + \alpha)$ then $\sin x< .6$.

Now $\sin x$ is symmetric and $\sin x = \sin (\pi - x)$ so for $x = \pi -\alpha$ we will get $\sin x = .6$ again but for the interval $x \in [\alpha, \pi - \alpha]$ we will have $\sin x \ge .6$.

Now for $\pi-\alpha < x < \pi$ we have $\sin x$ decreasing so $x \in (\pi-\alpha, \pi]$ we will have $\sin x < .6$.

And for all integer $k$ we get $((2k+1)\pi - \alpha, (2k+1)\pi]$ we will have $\sin x < .6$.

Now for $\pi < x < 2\pi$ we have $\sin x$ is negative and therefore less than $.6$. So we have $x\in (\pi, 2\pi)$ then $\sin x \le .6$.

Or in general, if $x \in ((2k-1) \pi, 2k\pi)$ then $\sin x \le .6$.

So in doing on $2\pi$ period we have $\sin x \le .6$ if $x \in [0, \alpha)$ or $x \in (\pi-\alpha, \pi]$ or $x \in (\pi, 2\pi)$. Combining those we get $\sin x < .6$ if $x\in [0, \alpha)$ or $(\pi-\alpha, 2\pi)$.

Putting in the general period we get $\sin x < .6$ if $x\in [2k\pi, 2k\pi + \alpha)$ or $x \in ((2k-1)\pi - \alpha, 2k \pi)$.

Combining we get $\sin x < .6$ if $x \in ((2k-1)\pi - \alpha, 2k\pi + \alpha)$

My solution is x < arcsin(0.6)+2kπ OR x<π−arcsin(0.6)+2kπ

Well, you aren't restricting to intervals. It works if $x < \arcsin .6$ but not if $x > \arcsin .6$. You can't just start with $x < \alpha$ and then just add $2k\pi$ to it. That would make you also insculde everything between $\alpha$ and $2k\pi$ and you know its not true for all of those values!

And as $k $ can be any integer then every value $x < \arcsin .6 + 2\pi k $ for some $k$!!!

When you (didn't) say $x < \arcsin .6$ it was understood you were restricting yourself to $-\frac \pi 2 \le x < \arcsin .6$. and when you (should have) said $x > \pi - \arcsin .6$ it was underston you were restricting yourself to $\pi - \arcsin .6 < x < 2\pi$.

So if you add $2k\pi$ to these you get $2k\pi -\frac \pi 2 \le x < \arcsin. 6 + 2k\pi$. And $2k\pi + \pi - \arcsin .6 < x < 2\pi + 2k\pi$. If we adjust and combine those we get $(2k-1)\pi -\arcsin .6 < x < 2k\pi + \arcsin .6$.

[And we DON'T get the result if $2k\pi + \arcsin. 6 \le x \le (2k_1)\pi - \arcsin .6$]

Answer 2

Look at the definition of $\arcsin$ function. The return value is in the interval $[-\pi/2,\pi/2]$. So the first thing is in this interval you get $$-\pi/2\le x\le\arcsin 0.6$$ You can write this as $$x\in[-\pi/2,\arcsin(0.6)]$$ You can now translate by $2k\pi$ and use the union of these intervals $$x\in\bigcup_{k\in\mathbb Z}[-\pi/2+2k\pi,\arcsin(0.6)+2k\pi]$$ Note however that you also have solutions in the $[\pi/2,3\pi/2]$ or eqivalent intervals.