This should be an easy exercise but I am stuck somehow in the definition of the equilibrium point:
Consider that the energy $V(u)$ on every point of a cement pile $u$ away from the stable state, satisfy the equation: $$\frac{dV}{du}=kL^2\sin(2u)-2PL\sin(u),$$ where $k,L,P$ are positive constants. Find all equilibrium points, namely the values of $V$ that correspond to distances $u$, where $V$ is constant.
But if we consider the function $f(u)=kL^2\sin(2u)-2PL\sin(u)$, I do not see how it is possible $f(u)=0$, for every $u\in(-\frac{\pi}{2},\frac{\pi}{2})$.
Later mark. I post also a figure just in case this makes the situation more clear...
To find critical points, you need values of $u$ such that $V'(u) = 0$ $$V'=a\sin 2u -b\sin u = 0$$ Therefore, $$2a\sin u \cos u - b\sin u = 0$$ So, either $$\sin u = 0$$ or $$\cos u = \frac{b}{2a}=\frac{P}{kL}$$
Can you take it from here?