why am I not finding pi as an extraneous solution? $2\sin^2(x/2)-3\sin(x/2)+1=0$

Question

$2\sin^2(x/2)-3\sin(x/2)+1=0$

I have solved this equation for $x = \pi, \pi/3, and 5\pi/3$. However, after graphing this equation on desmos, I noticed that there were only 2 solutions in the range of $x = [0, 2\pi)$. Why am I not getting pi as extraneous when I plug it back into the equation?

Answer 1

Maybe you plotted it wrong. There are three solutions in the interval $[0,2\pi)$, namely the ones you found. $\pi$ is not extraneous.

Answer 2

It is $$2t^2-3t+1=0$$ so $$t^2-\frac{3}{2}t+\frac{1}{2}=0$$ by the quadratic formula we get $$t_{1,2}=\frac{3}{4}\pm\sqrt{\frac{9}{16}-\frac{8}{16}}$$