Solve $\lfloor{\sin x}\rfloor+\lfloor{\cos x}\rfloor=2^{1-|\sin x|}$ 
Solve $$\lfloor{\sin x}\rfloor+\lfloor{\cos x}\rfloor=2^{1-|\sin x|}$$

Now, what I did notice is that $\lfloor{\sin x}\rfloor$ can only have three different values, and the same goes for $\lfloor{\cos x}\rfloor$. Moreover, if either is $1$ then the other is $0$, and $|\sin x|$ can only have one value. I am still not sure what are the solutions. Any help will be appreciated!
 A: You have $1 = 2^0 \le 2^{1-|\sin x|} \le 2^1 = 2$
$$\begin{array}{c|c|c} & \lfloor \sin x \rfloor & \lfloor \cos x \rfloor \\ \hline x = 0 & 0 & 1 \\ 0 < x < \dfrac{\pi}{2} & 0 & 0 \\ x = \dfrac{\pi}{2} & 1 & 0 \\ \dfrac{\pi}{2} < x \le \pi & 0 & -1 \\ \pi < x < \dfrac{3\pi}{2} & -1 & -1 \\ \dfrac{3\pi}{2} \le x < 2\pi & -1 & 0 \end{array}$$
So, the LHS intersects the possible range of the RHS at $x=0$ or $x=\dfrac{\pi}{2}$. So, we need only try those two values:
$$2^{1-|\sin 0|} = 2^1 = 2 \neq 1 \\ 2^{1-\left|\sin \tfrac{\pi}{2}\right|} = 2^0 = 1 = \left\lfloor \sin \dfrac{\pi}{2} \right\rfloor + \left\lfloor \cos \dfrac{\pi}{2}\right\rfloor$$
Thus, the solution is $x = \dfrac{\pi}{2}+2\pi n, n\in \mathbb{Z}$.
A: Observe that the right hand side $>0$ for real $x$
Now if $\sin x,\cos x<1$
$f(x)=\lfloor\sin x\rfloor+\lfloor\cos x\rfloor\le0$
So, we need at least one of them $=1$
Clearly $\sin x=1$ is a solution unlike $\cos x=1$
Also $f(x)\le\cos x+\sin x\le2$
So, $1-|\sin x|\le\dfrac12\iff|\sin x|\ge\frac12\implies\sin x\ne0\implies\cos x\ne\pm1$
