I'll give you some pointers to solve the exercise without complex calculations.
First you have to assume that the attenuation due to distance is none, i.e. the signal copies have the exact same amplitude.
Then, you can derive the wavelength from the frequency, that you can get in the signal formula (10^6*pi*t).
Then you know that when the signals reach the receiver with a certain phase difference, they cancel (sum is zero). What is this phase difference?
Then you can easily translate the phase difference to a distance, because you know the wavelength. And then you can say that at a certain position x, the difference between d1 and d2 will be exactly such that the signals cancel out. So you can say that at that point you will have deep fading.
A note on your solution: you say that you take the phasors of the signals to calculate the amplitude, then you just consider their rotational component, i.e. the phase. Mind that you're just analyzing their phase difference, and not their amplitude.
Also, in the equation that you feed to Alpha, I don't see x or any other variable, just the phase of the two signals that are fixed anyway. You can solve it numerically if you take:
$$
d_1 = d_2 + \Delta \phi
$$
With \$\Delta\phi\$ being the difference in phase that results in cancellation.
Every bit of wire has some equivalent series inductance. There is always a magnetic field formed looping around the current whenever current flows. The inductance isn't much for a straight wire, which is why inductors are made with coils of wire so this effect can be concentrated and magnified.
At high frequencies, these small inductances can become significant. One way to think of a inductor is as a frequency-variable resistor. That "resistance" goes up as the frequency goes up, so at some point that wire isn't a wire anymore for the purposes of your circuit.
Another issue is to minimize current loop area. Current flowing in a loop will radiate. That's what loop antennas are all about. However, loop antennas are very inefficient when the loop size is small compared to the wavelength. Since you don't want all the various current loops in your circuit to radiate or to pick up external radiation, you want to make them all as small as possible.
Best Answer
The answer is diffraction, and the fact that it takes larger objects to block the longer wavelengths.
1 MHz, which is in the middle of the commercial AM band, has a wavelength of 300 m. In contrast, 100 MHz, which is in the middle of the commercial FM band, has a wavelength of only 3 m.
300 m is large enough so that the waves can diffract around something the size of a typical house, for example. However, the house is much larger than 3 m, so it will largely block 100 MHz signal, assuming it is made of a material that blocks such frequencies. 1 MHz is much more able to "fill in" around house-size objects. At 100 MHz you get a lot more local dead zones and hot spots.
The real difference in propagation distance on the surface of the earth is due to the curvature and roughness of the earth. 300 m waves are able to refract around hills and the general curvature, whereas 3 m waves aren't. The smaller waves are more "line of sight" than the larger ones.
Of course there is still a huge difference between waves used for sight, around 500 nm, than those for commercial FM, around 3 m. The term "line of sight" for 3 m is therefore a bit misleading, the but the effect relative to 300 m waves is quite real. You can still pick up a 100 MHz station even with the antenna being a bit below the horizon while the visible beacon on the top of the antenna is completely blocked. But 3 m waves will get attenuated more quickly than 300 m waves as the transmitter gets further below the horizon.
Bouncing off the ionosphere is NOT the issue in most cases. It is true that 300 m waves can bounce off the ionosphere under the right conditions. This does allow picking up these stations significantly outside their usual broadcast area. However, ionospheric bounce is not why the usual broadcast area is larger in the first place.