Where did your "principal domain" come from? I think it would more likely be .
Then knowing that you should be able to find four solutions.
C)
Principal domain:
$-\dfrac{\pi}{2}\leq 2x^2+x-1\leq \dfrac{\pi}{2}$
$-1.41099 \leq x \leq 0.91099$
Next:
$\sin(2x^2+x-1)=\dfrac{2}{5}$
$2x^2+x-1=\arcsin(\dfrac{2}{5})$
$x_1=-1.1265; x_2=0.62650$
Since $\sin(\theta)=-\sin(\theta)+\pi$ we get:
$\sin(-(2x^2+x-1)+\pi)=\dfrac{2}{5}$
$-2x^2-x+1+\pi=\arcsin(\dfrac{2}{5})$
Hence:
$x_3=1.1384; x_4=-1.6384$
Although, I wouldn't be able to find any more values of x by the methods I have been introduced to, since the functions is aperiodic.
By the same method I can find two values of x in D), but that's it. Although, I think they want me to do something else, rather than values for x.
P.S. Also, I haven't been introduced to solving trig equations with aperiodic functions, so it must be that they want to stretch me a little. So the tasks are probably rigged somehow, like in the case of C) I can find exactly four values with the methods I've been shown (well, as I understand), and that's what they want from me in the problem.
So, I guess it must be similar with the D) task, but I can't figure it out.
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Great source of information, Cheers - Karol