Answer
Unstable at $\theta=34.6^{\circ}$; stable at $\theta=145^{\circ}$
Work Step by Step
We can determine the required positions for equilibrium and the stability at these positions as follows:
As given that, $V=12sin2\theta+15cos\theta$
But for equilibrium, $\frac{dV}{d\theta}=0$
$\implies \frac{d(12sin2\theta+15cos\theta)}{d\theta}= 24cos2\theta-15sin\theta=0$eq(1)
$\implies cos2\theta=\frac{5}{8}sin\theta$
This simplifies to:
$\theta_1=34.6^{\circ}$ and $\theta_2=145^{\circ}$
To check the stability at the above two positions, we take the second derivative of eq(1)
$\frac{d^2(12sin2\theta+15cos\theta)}{d\theta^2}=-48sin2\theta-15cos\theta$
Now, we plug in the values of $\theta_1$ and $\theta_2$ into the above equation
For $\theta_1=34.6^{\circ}$
$\implies \frac{d^2V}{d\theta^2}=-48sin(2(34.6^{\circ}))-15cos 34.6^{\circ}=-57.22\lt 0$
which means that it is in unstable equilibrium at this position.
For $\theta_2=145^{\circ}$
$\frac{d^2V}{d\theta^2}=-48sin(2(145^{\circ}))-15cos145^{\circ}=56.21\gt 0$
which means, there is a stable equilibrium at this position.
