$\Lambda$ for which $w(k)= k^3+8\Lambda \sin(k)$ has a minimum on x-axis. Given equation 
$$\begin{array}{cc}
\omega(k)= k^3+8\Lambda \sin(k) & \textrm{ with } k>0, \Lambda>0
\end{array}$$
Clearly $\omega(k)$ has a minimum for $k \approx 4$ which i will call $k_{min}$.
 For greater $\Lambda$ the minimum appears at a smaller $\omega$ value. 
Is there a way to determine analytically an explicit form of $\Lambda_c$ for which 
$\frac{d\omega}{dk}(k_{min})=0 $ and $\omega (k_{min})=0$ ?
 A: Assume that $\omega(k)=\omega'(k)=0$ then $a\sin k+k^3=a\cos k+3k^2=0$ with $a=8\Lambda$ hence $a^2=(a\cos k)^2+(a\sin k)^2=(3k^2)^2+(k^3)^2$. Solving the equation $x^3+9x^2=a^2$ for $x\gt0$ yields some explicit unique solution $x(a)$ (for example, using Cardano's formula for cubics) hence $a$ must solve $a\cos\sqrt{x(a)}+3x(a)=0$. Not sure one can make this more explicit...
A: There is not much hope for a closed form solution.
Consider the two equations
$$3k^2+8\Lambda\cos k =0 \\
k^3+8\Lambda\sin k=0.$$
Dividing them (and excluding $k=0$) we obtain
$$\tan k = \frac{\sin k}{\cos k} = \frac{-k^3}{-3k^2}\qquad\Rightarrow\qquad \tan k=\frac{k}{3}\qquad\text{(1)}$$
Adding them we have
$$(8\Lambda)^2(\sin^2 k +\cos^2 k) = (-3k^2)^2 + (-k^3)^2$$ and as $\sin^2 k+\cos^2 k =1$ we obtain 
$$\Lambda = \frac{k^2\sqrt{k^2+1}}{8}.\qquad\text{(2)}$$
To answer your question you need to solve (1) for $k$ and then substitute it into (2). However, because there is no known analytical solution for (1), there is also none for your problem (otherwise putting this hypothetical solution into (2) and inversion would lead to an analytical solution for (1)). 
The best you can do is solve (1) numerically or approximately based on additional assumptions, e.g. a perturbative expansion around $k\approx 4$.
