Again.. it's me. :)
So Let's go...
With the help of marginal law mentioned in my previous answer, we can progress.
The obvious place to start is ...
E12-G13 where an L-tromino is confirmed.
Then if we move to the top right cell
we see that no triangle fits there. And since K1 also has to be empty, so is L1. It's an I-tromino then! We can fill some cells.
Now let's try this.
If this happens, H1 and G2 will be empty cells, but then it causes a tetromino. Thus it shows that H1 has to be filled.
We can notice that J3
is currently surrounded on three sides, therefore it must also be an empty cell (remember this logic). It says that the adjacent cell also has to be empty. Hence we can complete another L-tromino (shown above).
Logic reveals that,
J3 cannot extend to an I-piece (Hint: consider K5). It must be a part of an L-piece. So H4 is cleared.
Also we can notice that if M6 is filled with a triangle,
we will get lost in an O-shape, which is not a tromino. Thus it is empty along with the cell beneath it.
Now consider the region I6-L8.
It can contain an L-tromino, and there are two possibilies. We can try them out
Now we see the emptiness of
I7, along with its companion H7.
Then we can fill G7.
with a prediction on F6.
And then G8 too.
Now we can complete our very first step noting that there is only one possibility for F11. It is..
Then we can
almost completely fill the right half of the grid.
I am not going to explain it, since it consists simple logical and trial and error methods.
Then, the bottom left part.
We can also fill this by simple chasing.
Now, this was the part I was stuck in, until I realized that,
I can make a BIG L-tromino.
From here, the rest is pretty easy.
And our finalized grid is..
... filled with trominoes!