The maximum tension in the string of an oscillating pendulum is double the minimum tension. Find the angular amplitude.
Approach with more physics and less maths:
Let $\theta$ be the amplitude.
$T = mg\cos x + \dfrac{mv^2}{l}$ , where T is the tension and $\dfrac{mv^2}l$ is the centripetal force.
$x$ is the angle made at any instant.
Also,
Using energy conservation:
$\dfrac{1}{2}{mv^2}= mgl(\cos x - \cos \theta)$
Solving the two we get,
$T = 3mg \cos x- 2mg \cos\theta$
Thus, tension is maximum at $x = 0$ and minimum at $x = \pm \theta$
Using the given condition,
$\theta = \arccos\left(\dfrac{3}{4}\right)$
Approach with more of maths:
$T = mg\cos x + \dfrac{mv^2}{l}$
$v = l\omega$ and for simple harmonic motion $\omega^2 = \dfrac{g}{l}({\theta^2-x^2}) \implies T = mg (\cos x+ \theta^2 -x^2 )$
using the given condition again,
$1+\theta^2= 2\cos\theta$ which has solution $0.714$
But $\arccos(0.75)\ne 0.714$
Why do the two different approaches yield different answers?
In the small-angle approximation, the motion of a simple pendulum is approximated by simple harmonic motion and not all ranges.
Because in small ranges you can approximate the motion to be linear-like and the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement.
Your problem is not limited to small angles.
To be concise, you cannot assume your second assumption because it is not simple harmonic. So, your answers won't be the same, however, it is close to the exact solution. This is because the angle has not reached too high and the error is still tiny.