Find the intervals in which the function $f$ given by $f(x)=\frac{4 \sin x-2 x-x \cos x}{2+\cos x}$ is $(i)$ increasing $(ii)$ decreasing.
(A) Given $f(x)=\frac{4 \sin x-2 x-x \cos x}{2+\cos x}$.
Using the quotient rule,$f'(x) = \frac{(2+\cos x)(4 \cos x - 2 - \cos x + x \sin x) - (4 \sin x - 2 x - x \cos x)(-\sin x)}{(2+\cos x)^2}$.
Simplifying the numerator: $(2+\cos x)(3 \cos x - 2 + x \sin x) + \sin x(4 \sin x - 2 x - x \cos x)$.
$= 6 \cos x - 4 + 2 x \sin x + 3 \cos^2 x - 2 \cos x + x \sin x \cos x + 4 \sin^2 x - 2 x \sin x - x \sin x \cos x$.
$= 4 \cos x - 4 + 3 \cos^2 x + 4 \sin^2 x = 4 \cos x - 4 + 3 \cos^2 x + 4(1 - \cos^2 x) = 4 \cos x - \cos^2 x$.
Thus,$f'(x) = \frac{\cos x(4 - \cos x)}{(2+\cos x)^2}$.
Since $(2+\cos x)^2 > 0$ and $(4 - \cos x) > 0$ for all $x$,the sign of $f'(x)$ depends on $\cos x$.
Setting $f'(x) = 0$ gives $\cos x = 0$,so $x = \frac{\pi}{2}, \frac{3\pi}{2}$ in $(0, 2\pi)$.
$(i)$ $f'(x) > 0$ when $\cos x > 0$,which occurs in $(0, \frac{\pi}{2}) \cup (\frac{3\pi}{2}, 2\pi)$.
$(ii)$ $f'(x) < 0$ when $\cos x < 0$,which occurs in $(\frac{\pi}{2}, \frac{3\pi}{2})$.
Using the quotient rule,$f'(x) = \frac{(2+\cos x)(4 \cos x - 2 - \cos x + x \sin x) - (4 \sin x - 2 x - x \cos x)(-\sin x)}{(2+\cos x)^2}$.
Simplifying the numerator: $(2+\cos x)(3 \cos x - 2 + x \sin x) + \sin x(4 \sin x - 2 x - x \cos x)$.
$= 6 \cos x - 4 + 2 x \sin x + 3 \cos^2 x - 2 \cos x + x \sin x \cos x + 4 \sin^2 x - 2 x \sin x - x \sin x \cos x$.
$= 4 \cos x - 4 + 3 \cos^2 x + 4 \sin^2 x = 4 \cos x - 4 + 3 \cos^2 x + 4(1 - \cos^2 x) = 4 \cos x - \cos^2 x$.
Thus,$f'(x) = \frac{\cos x(4 - \cos x)}{(2+\cos x)^2}$.
Since $(2+\cos x)^2 > 0$ and $(4 - \cos x) > 0$ for all $x$,the sign of $f'(x)$ depends on $\cos x$.
Setting $f'(x) = 0$ gives $\cos x = 0$,so $x = \frac{\pi}{2}, \frac{3\pi}{2}$ in $(0, 2\pi)$.
$(i)$ $f'(x) > 0$ when $\cos x > 0$,which occurs in $(0, \frac{\pi}{2}) \cup (\frac{3\pi}{2}, 2\pi)$.
$(ii)$ $f'(x) < 0$ when $\cos x < 0$,which occurs in $(\frac{\pi}{2}, \frac{3\pi}{2})$.
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