Answer
$\theta=38.7^{\circ}$ unstable, $\theta=90^{\circ}$ stable, $\theta=141^{\circ}$ unstable
Work Step by Step
We can determine the required equilibrium positions and stability at these positions as follows:
As given that, $V=10cos2\theta+25sin\theta$
At equilibrium, $\frac{dV}{d\theta}=0$
$\implies \frac{d(10cos2\theta+25sin\theta)}{d\theta}=-20sin2\theta+25cos\theta=0$eq(1)
This simplifies to:
$\theta_1=38.7^{\circ}$
$\theta_2=90^{\circ}$
and $\theta_3=141^{\circ}$
Now, we take the second derivative of eq(1) and plug in these values, one-by-one to obtain:
For $\theta_1=38.7^{\circ}$ $\frac{d^2V}{d\theta^2}=-40cos(2(38.7^{\circ}))-25sin(38.7^{\circ})=-24.375\lt 0$
which means unstable equilibrium.
For $\theta_2=90^{\circ}$ $\frac{d^2V}{d\theta^2}=-40cos(2(90))-25sin90=15\gt 0$
which means stable equilibrium.
For $\theta_3=141^{\circ}$ $\frac{d^2V}{d^2\theta}=-40cos(2(141))-25sin141=-24.375\lt 0$
which means unstable equilibrium.
