If $$\lfloor 5\sin x\rfloor+\lfloor \cos x\rfloor+6=0$$ then the solution set of range of
$$f(x)=\sqrt{3}\cos x+\sin x$$
I tried solving this question by doing the following
$5\sin x$ ranges from $[-5,5]$
So $\lfloor 5 \sin x\rfloor$ has the range $-5,-4,-3,-2,-1,1,2,3,4,5$.
Similar logic for $\lfloor\cos x\rfloor$;
$-5$ and $-1$ satisfies $\lfloor 5\sin x \rfloor+ \lfloor\cos x\rfloor+6=0$
But substituting these values in the second equation spent yield me the answer
But seems like I'm totally wrong as the correct answer would be $$\left(\frac{3×\sqrt3+4}{5},-1\right)$$
I think Im missing something conceptual here.
Could anyone help me out?
Note :
$\lfloor \cdot \rfloor$ indicates floor function
Best Answer
I think you are misinterpreting the question (Its wording is a bit weird imo). I would word it like this
Here is an image showing the situation for the interval $[0,2\pi]$:
In this picture, we have:
And you can solve it as follows:
Step 1: Determine the possible values of $x$: First we assume that $x\in[0,2\pi]$ because everything is periodic.
We need that $\sin x< -\frac45$ and $\cos x< 0$, because then $\lfloor 5\sin x\rfloor=-5$ and $\lfloor \cos x\rfloor=-1$ so the equation is satisfied. Clearly, $\cos x< 0$ gives $\frac\pi2< x<\frac{3\pi}2$ (The orange region). You can also solve $\sin x<-\frac45$ to obtain $2\pi-\sin^{-1}\frac45< x<\pi+\sin^{-1}\frac45$ (the red region).
So we find that $x$ is in the interval $\left(2\pi-\sin^{-1}\frac45,\frac{3\pi}2\right)$ (The green region).
Step 2: Find the values that $f$ takes in this interval. First we look at the value of $f$ at the boundary points $2\pi-\sin^{-1}\frac45$ and $\frac{3\pi}2$. We have that $f\left(2\pi-\sin^{-1}\frac45\right) = -\sqrt 3\sqrt{1-\left(-\frac45\right)^2}-\frac45=-\frac{3\sqrt 3+4}5$ and $f(\frac{3\pi}2)=-1$, so we at least know that the range contains these values.
Now we have to see if the function $f$ has a minimum or maximum in this interval. It does not (we see this in the image), so we are done.