1. Interpret the equation

   The given equation is $\cos^4\theta + \sin^4\theta + \lambda = 0.$

   For real solutions in $\theta$, we need $\cos^4\theta + \sin^4\theta = -\lambda$ to be attainable for some real $\theta$.

2. Find the range of $\cos^4\theta + \sin^4\theta$

   Use the identity $a^2+b^2=(a+b)^2-2ab.$

   Taking $a=\cos^2\theta$ and $b=\sin^2\theta$, we get $\cos^4\theta + \sin^4\theta = (\cos^2\theta+\sin^2\theta)^2 - 2\sin^2\theta\cos^2\theta.$

   Since $\cos^2\theta+\sin^2\theta=1,$ this becomes $\cos^4\theta + \sin^4\theta = 1 - 2\sin^2\theta\cos^2\theta.$

   Also, $\sin^2 2\theta = 4\sin^2\theta\cos^2\theta,$ so $2\sin^2\theta\cos^2\theta = \frac{1}{2}\sin^2 2\theta.$

   Therefore, $\cos^4\theta + \sin^4\theta = 1 - \frac{1}{2}\sin^2 2\theta.$

3. Determine its minimum and maximum

   Since $0 \le \sin^2 2\theta \le 1,$ we get $1 - \frac{1}{2} \le \cos^4\theta + \sin^4\theta \le 1.$

   Hence, $\frac{1}{2} \le \cos^4\theta + \sin^4\theta \le 1.$

4. Translate this to the condition on $\lambda$

   From $\cos^4\theta + \sin^4\theta + \lambda = 0,$ we have $-\lambda \in \left[\frac{1}{2},1\right].$

   Multiplying by $-1$ reverses the interval: $\lambda \in [-1,-\tfrac12].$

5. Match with the options

   The correct interval is $[-1,-\tfrac12],$ which corresponds to Option D.

Final Answer: $\boxed{D}$