I'm stuck on this quadratic problem: $$\sin^4 x + 2(1 + k)\sin^2 x- k - 2 = 0$$ $$k = ?$$ I substituted $t = \sin^2(x)$ this gave me $t^2 + 2(1 + k)t - k - 2 = 0$ and than I figured out that the discriminant is $k^2 + 3k + 3 \ge 0$ which is true for $k \in \mathbb{R}$. The other thing I figured out is that $t \in [0, 1]$ and this is where I don't know how to continue. The answer is supposed to be $k \in [-2, 1]$.
Edit: Forgot to mention that I'm interested in real number solutions only.
On
Note that since $0$ and $1$ enclose both the roots, and the given quadratic expression has positive leading coefficient, so the quadratic expression should take non-negative sign at both $0$ and $1$, or at least one of them should be the root. So if $f(t)$ denotes this quadratic expression, we have $f(0)\ge0$, and $f(1)\ge0$. Hence $$(-k-2)\ge0\qquad\text{and}\qquad k+1\ge0$$ which is impossible, because $k+1\ge 0\implies k+2\ge 1\implies -k-2\le -1$. Hence no such $k$ exists.
Express the equation as
$$k = \frac{2-2\sin^2 x - \sin^4 x }{2 \sin^2 x -1} = \frac14 \left( \cos 2x - \frac3{\cos 2x}-6\right) $$ For $x$ in the real domain, $k$ has the range $k\notin (-2,-1)$.