Hint

Let $y=\pi\log x.$ Then $\sin y + \cos y = 1.$ Squaring, we get $2\sin y\cos y = 0,$ hence $\sin y = 0$ or $\cos y = 0.$ Thus $y = k\pi/2, k \in \mathbb{Z}.$ However, when $k \equiv 2, 3 \mod 4,$ the solution is extraneous. Thus $x=e^{k/2}, k \in \mathbb{Z}$ where $k \equiv 0, 1 \mod 4$ is the vinyl solution.

twinbrian wrote:

Hint

This is not true. $x=1$ is a counter example.

AlastorMoody wrote: Solve the equation,$$ \sin (\pi \log x) + \cos (\pi \log x) = 1$$ Hello, use that $$\sin(x)+\cos(x)=\sqrt{2}\sin\left(\frac{\pi}{4}+x\right)$$ Sonnhard.

Squaring, we see that $$\sin(\pi \log x) + \cos(\pi \log x) = 1 \implies $$$$\sin^2(\pi \log x) + 2\sin(\pi \log x)\cos(\pi \log x) + \cos^2(\pi \log x) = 1 \implies$$$$1 + 2\sin(\pi \log x)\cos(\pi \log x) = 1 \implies$$$$2\sin(\pi \log x)\cos(\pi \log x) = 0 \implies$$$$\sin(2\pi \log x) = 0.$$From here, it becomes evident that $\log x = \frac{1}{2} + 2k,2k$, where $k$ is an integer. Hence, $x=e^{k/2}$, where $k \equiv 0, 1 \mod 4$.

We can square this to get that either sin or cos of $\pi \text{log}x$ is $0$. Then easily finish.