Question
Find the intervals in which the function $f(x) = \frac{4\sin x}{2 + \sin x} -x, 0\leq x\geq2\pi$ is strictly increasing or strictly decreasing.

Answer

$\text{y} = \frac{4\sin\text{x}}{2 + \cos \text{x}} - \text{x, x} \in [0, 2\pi] $
$\frac{\text{dy}}{\text{dx}} = \frac{(2 + \cos\text{x)}4\cos\text{x} - 4\sin\text{x} (- \sin \text{x)}}{(2 + \cos\text{x})^{2}} - 1$
$\frac{\text{dy}}{\text{dx}} = \frac{\cos\text{x}(4 - \cos\text{x})}{(2 + \cos\text{x})^{2}}$
$\text{f(x) is strictly increasing for f'(x) > 0}$
$\text{i.e.}\ \ \ \cos \text{x} > 0 \Rightarrow\text{x} \in\bigg[0, \frac{\pi}{2}\bigg)\cup\bigg(\frac{3\pi}{2}, 2\pi\bigg]$
$\text{and f(x) is strictly decreasing for f'(x) < 0}$
$\text{i.e}\cos\text{x}<0\Rightarrow\text{x}\in\bigg(\frac{\pi}{2},\frac{3\pi}{2}\bigg)$

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