solving trigonometry Solve $16 \cos^2 x + 6 \sin x = 17$ for $0 < x < 2\pi$
Steps:
$16(1-\sin^2x)+6\sin x = 17$
$16\sin^2x-6\sin x + 1 = 0$
let $y = \sin x$
$16y^2 - 6y + 1 = 0$
I was not able to solve this quadratic equation. It has no real roots. What was done incorrectly?

As noticed your derivation is fine, the quadratic equation $16y^2 - 6y + 1 = 0$ has not real roots indeed $f(y)=16y^2 - 6y + 1$ represents a parabola that opens upward with vertex at $\left(\frac 3{16},\frac 7{16}\right)$.
As an alternative, we have that
$$16 \sin^2 x - 6 \sin x + 1=0 \iff 16 \sin^2 x +1= 6 \sin x $$
which can't have solutions for $\sin x \le 0$ and for $\sin x >0$ by AM-GM
$$16 \sin^2 x +1\ge 8 \sin x>6 \sin x$$
