Describe $f(S)$ where S is the solution set of equation $\lfloor5\sin x\rfloor+\lfloor\cos x\rfloor+6=0$

Question

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

Answer 1

I think you are misinterpreting the question (Its wording is a bit weird imo). I would word it like this

When $x$ satisfies $$\lfloor 5\sin x\rfloor+\lfloor \cos x\rfloor+6=0,$$ then what values can the function $$f(x)=\sqrt 3\cos x+\sin x$$ obtain for these $x$.

Here is an image showing the situation for the interval $[0,2\pi]$:

In this picture, we have:

• Orange region is where $\lfloor\cos x\rfloor = -1$.
• Red region is where $\lfloor 5\sin x\rfloor = -5$.
• Green region is where $\lfloor 5\sin x\rfloor+\lfloor\cos x\rfloor+6=0$.

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.

Answer 2

Since $-1\leqslant \lfloor\cos x\rfloor$ always holds, we have to have $-1\leqslant -6-\lfloor 5\sin x\rfloor$, i.e. $\lfloor 5\sin x\rfloor \leqslant -5$.

Since $-5\leqslant \lfloor 5\sin x\rfloor$ always holds, we have to have $$\lfloor 5\sin x\rfloor =-5\iff -1\leqslant \sin x\lt -\frac 45\tag1$$ and $$\lfloor\cos x\rfloor=-1\iff -1\leqslant \cos x\lt 0\tag2$$ These are sufficient.

Solving $(1)(2)$ for $0\leqslant x\lt 2\pi$, we get $\alpha\lt x\lt\dfrac{3\pi}{2}$ where $\sin\alpha=-\dfrac 45$ with $\dfrac{7}{6}\pi\lt \alpha\lt\dfrac{3}{2}\pi$ since $\sin\alpha=-\dfrac 45\lt -\dfrac 12=\sin\dfrac{7\pi}{6}$.

We have $$f(x)=\sqrt 3\cos x+\sin x=2\sin\bigg(x+\frac{\pi}{3}\bigg)$$ and $$\bigg(\frac{3\pi}{2}\lt\bigg)\alpha+\frac{\pi}{3}\lt x+\frac{\pi}{3}\lt\frac{11\pi}{6}\bigg(\lt 2\pi\bigg)$$ Since $y=\sin x$ is strictly increasing for $\dfrac{3\pi}{2}\lt x\lt 2\pi$, the range is $$2\sin\bigg(\alpha+\frac{\pi}{3}\bigg)\lt f(x)\lt 2\sin\bigg(\frac{11\pi}{6}\bigg)$$ i.e. $$2\bigg(\sin\alpha\cos\frac{\pi}{3}+\cos\alpha\sin\frac{\pi}{3}\bigg)\lt f(x)\lt -1$$ i.e. $$2\bigg(-\frac 45\cdot\frac 12-\sqrt{1-\bigg(-\frac 45\bigg)^2}\cdot\frac{\sqrt 3}{2}\bigg)\lt f(x)\lt -1$$ i.e. $$\color{red}{\frac{-4-3\sqrt 3}{5}\lt f(x)\lt -1}$$