Find the intervals in which the function $f:R \rightarrow R$, $f\left(x\right) = \dfrac{4 \sin x -2x - x \cos x}{2+\cos x}$ is increasing and in which it is decreasing.
$f\left(x\right) = \dfrac{4 \sin x -2x - x \cos x}{2 + \cos x}$
i.e. $f\left(x\right) = \dfrac{4\sin x}{2+\cos x} - \dfrac{x \left(2+\cos x\right)}{2+\cos x}$
i.e. $f\left(x\right) = \dfrac{4\sin x}{2+\cos x} - x$
$\therefore$ $f'\left(x\right) = \dfrac{\left(2+\cos x\right)\times 4 \cos x -4 \sin x \times \left(-\sin x\right)}{\left(2+\cos x\right)^2} -1$
i.e. $f'\left(x\right) = \dfrac{8 \cos x + 4 \cos ^2 x + 4 \sin ^2 x}{\left(2+\cos x\right)^2}-1$
i.e. $f'\left(x\right) = \dfrac{8 \cos x + 4 -4 - 4\cos x -\cos^2 x}{\left(2+\cos x\right)^2}$
i.e. $f'\left(x\right) = \dfrac{4 \cos x - \cos^2 x}{\left(2+\cos x\right)^2}$
i.e. $f'\left(x\right) = \dfrac{\cos x \left(4- \cos x\right)}{\left(2+\cos x\right)^2}$ $\;\; \cdots$ (1)
In equation (1), $\left(2+\cos x\right)^2 > 0 \; \forall \; x$ $\;\; \cdots$ (2)
Also, $4 - \cos x > 0 \; \forall \; x$ since $\cos x$ varies between $\pm 1$ $\;\; \cdots$ (3)
Case (i): $\mathbf{f\left(x\right)}$ is increasing:
$f\left(x\right)$ is increasing when $f'\left(x\right) > 0$
$\therefore$ From equation (1), for $f'\left(x\right) > 0$, $\cos x > 0$
Now, $\cos x > 0$ i.e. positive in the First and Fourth quadrants.
$\therefore$ $\cos x > 0$ $\implies$ $0 < x < \dfrac{\pi}{2}$ (First Quadrant) and $\dfrac{3\pi}{2} < x < 2\pi$ (Fourth Quadrant)
$\therefore$ Generalizing,
$\cos x > 0$ $\implies$ $2k\pi < x < \left(4k+1\right)\dfrac{\pi}{2}$ (First Quadrant) and $\left(4k+3\right) \dfrac{\pi}{2} < x < \left(2k+2\right)\pi$ (Fourth Quadrant), $k \in Z$
$\therefore$ $f\left(x\right)$ is increasing in the interval $\left(2k\pi, \left(4k+1\right)\dfrac{\pi}{2}\right)$ and $\left(\left(4k+3\right)\dfrac{\pi}{2}, \left(2k+2\right)\pi\right)$, $k \in Z$
Case (ii): $\mathbf{f\left(x\right)}$ is decreasing:
$f\left(x\right)$ is decreasing when $f'\left(x\right) < 0$
$\therefore$ In view of equations (2) and (3), we have from equation (1), for $f'\left(x\right) < 0$, $\cos x < 0$
Now, $\cos x < 0$ i.e. negative in the Second and Third quadrants.
$\therefore$ $\cos x < 0$ $\implies$ $\dfrac{\pi}{2} < x < \pi$ (Second Quadrant) and $\pi < x < \dfrac{3\pi}{2}$ (Third Quadrant)
$\therefore$ Generalizing,
$\cos x < 0$ $\implies$ $\left(4k+1\right)\dfrac{\pi}{2} < x < \left(2k+1\right)\pi$ (Second Quadrant) and $\left(2k+1\right)\pi < x < \left(4k+3\right) \dfrac{\pi}{2}$ (Third Quadrant), $k \in Z$
$\therefore$ $f\left(x\right)$ is decreasing in the interval $\left(\left(4k+1\right)\dfrac{\pi}{2}, \left(2k+1\right)\pi \right)$ and $\left(\left(2k+1\right) \pi, \left(4k+3\right)\dfrac{\pi}{2}\right)$, $k \in Z$