Finding out the range of $g(x)=\sin x + \cos x$

Question

If $x$ is such that $$\lfloor2\sin x\rfloor+\lfloor{\cos x}\rfloor=-3$$ for some $x \in [0,25]$ then the question is to find out the range of the function $$g(x)=\sin x+\cos x$$Here the function $g(x)$ assumes those values of $x$ which satisfies the above equation in the given interval

$\lfloor.\rfloor$ represents greatest integer function.

I tried rewriting the equation as $$2 \sin x+\cos x -2\lbrace(\sin x)\rbrace-\lbrace(\cos x)\rbrace=-3$$ which can be rewritten as $$g(x)=-3+2\lbrace(\sin x)\rbrace+\lbrace(\cos x)\rbrace-\sin x$$ I again rewrote it as $$g(x)=-3-\lfloor\sin x\rfloor+\lbrace(\sin x)\rbrace+\lbrace(\cos x)\rbrace$$ $\lfloor\sin x\rfloor=\lbrace 0,-1 \rbrace$.I tried to further find out the range of this but failed.Any ideas?Thanks.

Here $\lbrace.\rbrace$ represents fractional part.

Answer 1

Note that $\lfloor \cos x \rfloor $ only takes values $-1,0,1$.

Let $S = \{x \in [0,2 \pi] | \lfloor \cos x \rfloor + \lfloor 2\sin x \rfloor = -3 \}$.

If $\lfloor \cos x \rfloor = 0$, then we would need to have $\lfloor 2\sin x \rfloor = -3$ which is impossible.

If $\lfloor \cos x \rfloor = 1$, then we would need to have $\lfloor 2\sin x \rfloor = -4$ which is impossible.

Hence we must have $\lfloor \cos x \rfloor =-1$ and so

If $\lfloor \cos x \rfloor =-1$, then we must have $\lfloor 2\sin x \rfloor = -2$, and hence $x \in ({7 \over 6} \pi, {9 \over 6} \pi)$.

Hence $S = ({7 \over 6} \pi, {9 \over 6} \pi)$.

We note that $g$ has a (global) $\min$ at $x={5 \over 4} \pi$, and $g({5 \over 4} \pi) = -\sqrt{2}$.

Since $g$ is unimodal in $S$, we see that $g(S) = [-\sqrt{2}, -1)$ (since $\max(g({7 \over 6} \pi), g({9 \over 6} \pi)) = -1$).

Answer 2

Considering functions' periods, without loss of generality we can restrict to $x\in\left[ 0,\,2\pi\right)$.

The first insight you missed is that each integer part in the problem statement must minimise its value to achieve $-3$. We can then restate the constraint as $\sin x\in\left[ -1,\,-\frac{1}{2}\right),\,\cos x\in\left[ -\frac{\sqrt{3}}{2},\,0\right)$, the latter result tightened with $\sin^2 x+\cos^2 x=1$. The range of "legal" values is $S:=\left(\frac{7\pi}{6},\,\frac{3\pi}{2}\right)$.

The second insight you need is a restatement of the function we wish to extremise. We seek the range of $\sqrt{2}\sin\left( x+\frac{\pi}{4} \right)$ on $S$. We reach a minimum of $-\sqrt{2}$ at $x=\frac{5\pi}{4}$ and a supremum of $-1$ at $x=\frac{3\pi}{2}$ (since $\sin\frac{7\pi}{4}=-\frac{1}{\sqrt{2}}$), which is technically outside $S$. The range is thus $\left[ -\sqrt{2},\,-1\right)$.

Answer 3

Hint: Here is a visualization.

(Large version)

The red lines are the graph of $f(x) = \lfloor2\sin x\rfloor+\lfloor{\cos x}\rfloor$. The blue line is $h(x) = -3$. And the green line is $g(x) = \sin x+\cos x$.

The $x$ coordinates of the intersection of the graphs of $f$ and $h$, intersected with $[0,25]$ will determine a set $A$ of $x$ values. What you seek is $g(A)$.