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# Find the intervals on which $f$ is increasing or decreasing. $f(x)=x-2\sin x,[0 , 2\pi ]$

Note: This is part 4th of a 6 part question, split as 6 separate questions here.

Toolbox:
• (i) If $f'$ is positive on an open interval $I$. Then $f$ is strictly increasing on $I$
• (ii) If $f'$ is negative on an open interval $I$, then $f$ is strictly decreasing on $I$
$f(x)= x -2 \sin x \quad x \in[0,2\pi]$
Step 1:
$f(x)=1-2 \cos x$
$f'(x)=0=>1-2 \cos x =0=>\cos x =\large\frac{1}{2}$
Step 2:
$\therefore$ the values of $x \in [0,2 \pi]$ for which $\cos x =\large\frac{1}{2}$ are $\large\frac{\pi}{3}, \frac{5 \pi}{3}$
Step 3:
Interval $(0, \large\frac{\pi}{3})$
$f'(x)=1-2 \cos x$=> $-$ (since $\cos x > \large\frac{1}{2})$
Interval of inc /dec $f(x)$= $f(x)$ is decreasing in $[0,\large\frac{\pi}{3}]$
Interval $( \large\frac{\pi}{3},\frac{5\pi}{3})$
$f'(x)=1-2 \cos x$=> $-$ (since if positive $\cos x < \large\frac{1}{2}$ or negative)
Interval of inc /dec $f(x)$= $f(x)$ is increasing in $[\large\frac{\pi}{3},\frac{5 \pi}{3}]$
Interval $( \large\frac{5\pi}{3},$$2\pi) f'(x)=1-2 \cos x=> - (since \cos x > \large\frac{1}{2}) Interval of inc /dec f(x)= f(x) is decreasing in [\large\frac{5\pi}{3},$$2 \pi]$
$\therefore f(x)$ is decreasing in $[0,\large\frac{\pi}{3}] $$\cup [\large\frac{5 \pi}{3}$$,2 \pi]$ and increasing in $[\large\frac{\pi}{3},\frac{ 5\pi}{3}]$