In the string of an oscillating pendulum, the maximum tension is...

Let the angular amplitude be $\theta_0$.

The tension in the string at an angle $\theta$ with the vertical is given by $T = mg \cos \theta + \dfrac{mv^2}{l}$.

By conservation of energy, the velocity $v$ at angle $\theta$ is given by $\dfrac{1}{2} m v^2 = mg l (\cos \theta - \cos \theta_0)$, which gives $v^2 = 2gl (\cos \theta - \cos \theta_0)$.

Substituting $v^2$ into the tension equation, we get $T = mg \cos \theta + 2mg (\cos \theta - \cos \theta_0) = mg (3 \cos \theta - 2 \cos \theta_0)$.

The maximum tension occurs at the lowest point ($\theta = 0$):
T_\max = mg (3 - 2 \cos \theta_0)

The minimum tension occurs at the extreme position ($\theta = \theta_0$):
T_\min = mg (3 \cos \theta_0 - 2 \cos \theta_0) = mg \cos \theta_0

Given that T_\max = 2 T_\min:
$mg (3 - 2 \cos \theta_0) = 2 mg \cos \theta_0$

$3 - 2 \cos \theta_0 = 2 \cos \theta_0$

$4 \cos \theta_0 = 3$

$\cos \theta_0 = \dfrac{3}{4}$

$\theta_0 = \cos^{-1} \left(\dfrac{3}{4}\right)