Home / Let $t = \sin 2x$, $0 < t \leq 1$ (since $x \in (0, \pi/2)\Rightarrow 2x \in (0, \pi)$, and $t = 1$ at $x = \pi/4$)

Let $t = \sin 2x$, $0 < t \leq 1$ (since $x \in (0, \pi/2)\Rightarrow 2x \in (0, \pi)$, and $t = 1$ at $x = \pi/4$)

February 22, 2026

$Let $t = \sin 2x$, $0 < t \leq 1$ (since $x \in (0, \pi/2)\Rightarrow 2x \in (0, \pi)$, and $t = 1$ at $x = \pi/4$)$

an x + \cot x = rac{\sin x}{\cos x} + rac{\cos x}{\sin x} = rac{\sin^2 x + \cos^2 x}{\sin x \cos x} = rac{1}{\sin x \cos x} = rac{2}{\sin 2x}
Now the full expression:
+ \sin 2x + rac{4}{\sin^2 2x}
Define $f(t) = 1 + t + rac{4}{t^2}$, $0 < t \leq 1$
Derivative: $f'(t) = 1 - rac{8}{t^3}$
Set to zero: $1 = rac{8}{t^3} \Rightarrow t^3 = 8 \Rightarrow t = 2$, not in domain. So maximum occurs at endpoint $t = 1$