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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.

Quanto
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Vojtěch Kříbek
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  • Do you need both the roots to be in $[0,1]$? – Vishu Apr 09 '21 at 13:00
  • @Tavish, Yeah because $\sin^2x\in[0,1]:\forall:x\in\mathbb R$. – Martund Apr 09 '21 at 13:03
  • @Martund It could also be the case that one of the roots is in $[0,1]$, while the other is redundant. – Vishu Apr 09 '21 at 13:04
  • Oh @Tavish, I didn't think of that. – Martund Apr 09 '21 at 13:04
  • The question is very unclear as written: $k$ could be anything. – Vishu Apr 09 '21 at 13:08
  • @Tavish, look at my edited answer, $k$ cannot be anything lol. – Martund Apr 09 '21 at 13:13
  • @Martund It can be. Why do you assume this quadratic must have real solutions? Why do you assume $x\in\mathbb R$? – Vishu Apr 09 '21 at 13:15
  • @Tavish, because the discriminant is non-negative. – Martund Apr 09 '21 at 13:16
  • @Martund That only means $\sin^2 x$ can be any of two real values, not necessarily in $[0,1]$ because $\sin x$ may be say, $2i$. – Vishu Apr 09 '21 at 13:20
  • @Tavish, I strictly believe the OP is uninterested in complex values of sine function. It's a basic quadratic equations question. (Since the OP never used analysis, complex numbers, real analysis, or any such tag.) – Martund Apr 09 '21 at 13:28
  • @Martund That’s just a belief, you don’t know for sure. Although it is probable. My point is that more details should be added in the question. – Vishu Apr 09 '21 at 13:29

2 Answers2

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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.

Martund
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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 $k\notin (-2,-1)$.

Quanto
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