From what I've read, planets orbiting red dwarf stars would most likely be tidally locked. Under what circumstances might this change? For example, if a planet had been the recipient of some kind of large impact over the course of its existence? We theorize that in the Solar System, Venus's reverse orbit might have been caused by a big impact. If such a thing were to occur on a planet orbiting a red dwarf, could it spin on some kind of "normal" rotational period (e.g., 5-100 hour days).
1) A possibility to make a planet less tidally locked would be libration.
Luna, the Moon, has libration that makes it seem to wobble very slightly as it orbits the Earth.
Tidal locking means that the period that the Moon takes to revolve or orbit 360 degrees around the Earth is exactly the same as the period it takes the moon to rotate 360 degrees. So one side of the Moon always faces the Earth and one side always faces away from the Earth. This is because the average orbital speed and the average rotation speed of the Moon are identical.
But the speeds at any particilar moment are not exactly identical. The orbits of all astronomical objects are elliptical and thus deviate more or less from perfectly circular orbits. Thus objects speed up and slow down as they orbit other objects. So the Moon sometimes travels faster and sometime slower than it's average orbital speed. But the Moon cannot speed up or slow down its rotation, it has to always rotate at its average rotation speed.
So a total of 59 percent of the Lunar surface is visible from Earth, instead of fifty percent.
So if your planet orbited its sun in exactly 3 Earth days or 72 Earth hours, it would orbit at an average rate of 120 degrees per Earth day or 5 degrees per Earth hour. And if tidally locked it would rotate at exactly 120 degrees per Earth day or 5 degrees per Earth hour. And if the planet's orbit is highly eccentric the planet's variable orbital speed would sometimes be faster or slower than it's exact rotation rate, thus making for a lot of libration and for a lot more than 59 percent of the planet's surface to sometimes be in daylight.
In our solar system the planet Mercury has an orbital eccentricity of 0.2563 and is 1.5177 times farther from the Sun at Aphelion distance than at Perihelion distance. If Mercury was tidally locked to the Sun its libration would be 23.65 degrees due to its eccentric orbit.
Of course the same tidal forces that would tend to lock the planet's rotation would also tend to make its orbit more and more circular with time.
2) Maybe, like Mercury, its orbit and rotation might become locked in a 2:3 resonance.
Mercury has an orbital period or year that is 87.9691 Earth days long. It has a rotation period relative to distant stars (or sidereal day) 56.646 Earth days long. That means that it's solar day, it's day relative to the sun, is two Mercurian years, or about 176 Earth days long.
If another solar system had a much smaller scale version of Mercury's orbit and rotation, each might be, for example, one twentieth as long as that of Mercury.
The hypothetical planet could have an orbital period or year that is 4.398 Earth days long, with a rotational period period relative to distant stars (or sidereal day) of 2.8323 Earth days. That means that it's solar day, it's day relative to its sun, would two of its years, or about 8.796 Earth days long.
So divide those figures by three to get a year 1.442 Earth days long, a sidereal day 0.9441 Earth days long, and a solar day 2.932 Earth days, or 70.368 Earth hours, long.
Can a star have such a close habitable zone that a planet would have a year only 1.442 Earth days, or 34.608 Earth hours, long?
Here is a discussion of distances of habitable planets from their stars:
Apparently K3-137b has the shortest known year of any know exoplanet, 4.31 hours, orbiting a red dwarf - except that PSR J1719-1438 orbits a pulsar every 2.2 hours. But they are not listed as potentially habitable planets.
The list of known exoplanets in the conservative habitable zone includes ones with years 12.4, 9.2, 6.1, and 4.05 Earth days long. They orbit TRAPPIST-1, a M8V type star. I don't know if a M9V typed star could be enough dimmer to have a planet orbiting it in the habitable zone with a year as short as 1.442 Earth days.
For a solar day 100 Earth hours long, a habitable planet with Mercury-like resonance would have a year 2.0492 Earth days long, a sidereal day 1.341 Earth days, long and a solar day 4.0984 Earth days, or 98.3616 hours long.
It is possible that your planet could orbit in the habitable zone of a brown dwarf, intermediate between a planet and and star. Thus it might have a year and a solar day shorter than a planet orbiting even the dimmest red dwarf star.
But it seems very likely such a planet would be tidally locked.
I may point out that the apparent movement of the sun in Mercury's sky is sometimes very odd, and that different places on Mercury's equator will have significantly different temperatures during their noonday periods. If this planet planet orbits a red star in the habitable zone the tropics and the temperate zone would be divided by longitude as well as latitude.
Why does Mercury have a 2:3 orbit: spin resonance? Scientists are still coming up with ideas and computer simulations to explain it.
For example, some scientists have proposed that Mercury was once tidally locked to the Sun but a massive asteroid strike changed its rotation period, similar to what paltrysum suggested.
Other computer simulations suggest that a 2:3 resonance is a more natural situation and likely to occur a lot in extrasolar planets.
You should look for other questions and answers about habitable planets in 2:3 resonances.
There is also the suggestion that a habitable planet could actually be a habitable moon of giant planet in the habitable zone of a red star. Thus the habitable moon would be tidally locked to the planet instead of to the star and would have periods of daylight and darkness.
It should be noted that it has been calculated that the orbit of a moon will not be stable unless its month is one ninth or less the length of it's planet's year - the length of the planet's year should be at least 9 times the length of the moon's month. If it is desired that the day on the moon, equal in length to the moon's month, should be about 5 to 100 Earth hours, the length of the planet's year should be at least 45 to 900 Earth hours, or 1.875 to 37.5 Earth days, and it can be several times as long. Thus the length of the planet's year can be made to fit into the the possible length of a year in the habitable zone of a red dwarf star.
You should look for other questions and answers about habitable moons of giant planets in the habitable zones of red dwarf stars.
Are you talking about planets in the goldilocks zone?
If not, simply have the planet be further away from the star. The reason it is thought that planets in the goldilocks zone (the band around a sun where liquid water is possible) of a red dwarf would likely be tidally locked is because they'd have to be so close to their parent star, since red dwarfs are small and dim. If you remove this constraint, you can just move the planet away and it would be able to rotate freely.
In this case the planet wouldn't be terrestrial, but it wouldn't be tidally locked.
If so, and you are in fact talking about planets in the habitable zone, then one proposed solution (which I found on the astromony stackexchange), would be to have the planet be part of a double system with the other member either being a very large moon or even a fellow planet (depending on your definitions).
Systems like these exist in the real world. The Pluto–Charon system being one such example.
In this fashion both bodies would be tidally locked to each other (much as the Moon is to Earth), but not tidally locked to their star.