This can be solved in a far simpler way: -

[![enter image description here][1]][1]

So, to get the RMS value of the waveform, you square the individual parts (triangle and DC) to get power into a 1 Ω resistor, weight them with their duty cycle, add them together and finally, take the square root to get back to RMS voltage and lose the 1 Ω dependency.

 - For the triangle section, it's weighted power is \$\dfrac{4}{3}\times 7\div 15\$
 - For the DC part it's just \$4\times 8\div 15\$
 - Add them to get 2.755555
 - Take the square root to get 1.65998661307

[Proof of triangle waveform RMS](https://electronics.stackexchange.com/questions/499312/calculating-rms-of-simple-waveform/499343#499343): -

[![enter image description here][2]][2]


  [1]: https://i.sstatic.net/RLhpT.png
  [2]: https://i.sstatic.net/YKC40.png