Find the interval of monotonicity for x - |sin(2x)|

[sonic marsh]

$\text{Find the interval of monotonicity:}\
f(x) = x + \sin(2x)\
\text{To find the intervals of monotinicty, I calculated f'(x)}\
f'(x) = 1 - \frac{\sin(x)cos(x)}{|\sin(x)|} = \frac{|\sin(x)| - \sin(x)cos(x)}{|\sin(x)|} \
\text{To find the interval of monotonicity we need to find the negative and positive intervals of f'}\
\sin(x) > 0 \text{ , for } x\in (0 , \pi) \
\sin(x) < 0 \text{ , for } x\in (\pi ,2\pi) \
\cos(x) > 0 \text{ , for } x\in (0, \frac{\pi}{2}) \cup x\in (\frac{3\pi}{2}, 2\pi) \
\cos(x) < 0 \text{ , for } x\in (\frac{\pi}{2}, \frac{3\pi}{2}) \
\text{So we get that:}\
\sin(x) * \cos(x) > 0 \text{, for } x\in (0, \frac{\pi}{2}) \cup (\pi, \frac{3\pi}{2})\
\sin(x) * \cos(x) < 0 \text{, for }x\in (\frac{\pi}{2}, \pi) \cup (\frac{3\pi}{2}, 2\pi)\
\text{Where do I go from here? How do I find the interval where:}\
|\sin(x)| - \sin(x)\cos(x) < 0 \
\text{without access to a calculator?}$

[errant oak]

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[old owl]

Lonac

Lonac

[sonic marsh]

$\text{Find the interval of monotonicity:}\\
f(x) = x + |\sin(2x)|\\
\text{To find the intervals of monotinicty, I calculated f'(x)}\\
f'(x) = 1 + \frac{\sin(x)cos(x)}{|\sin(x)|} = \frac{|\sin(x)| + \sin(x)cos(x)}{|\sin(x)|} \\
\text{To find the interval of monotonicity we need to find the negative }\\text{and positive intervals of f'}\\
\sin(x) > 0 \text{ , for } x\in (0 , \pi) \\
\sin(x) < 0 \text{ , for } x\in (\pi ,2\pi) \\
\cos(x) > 0 \text{ , for } x\in (0, \frac{\pi}{2}) \cup x\in (\frac{3\pi}{2}, 2\pi) \\
\cos(x) < 0 \text{ , for } x\in (\frac{\pi}{2}, \frac{3\pi}{2}) \\
\text{So we get that:}\\
\sin(x) * \cos(x) > 0 \text{, for } x\in (0, \frac{\pi}{2}) \cup (\pi, \frac{3\pi}{2})\\
\sin(x) * \cos(x) < 0 \text{, for }x\in (\frac{\pi}{2}, \pi) \cup (\frac{3\pi}{2}, 2\pi)\\
\text{Where do I go from here? How do I find the interval where:}\\
|\sin(x)| + \sin(x)\cos(x) < 0 \\
\text{without access to a calculator? EDIT: corrected typo while in latex}$

[old owl]

Lonac

[strong marten]

|sin(2x)|=sqrt(sin^2(2x)) so deriving it gives 2cos(2x)sin(2x)/2sqrt(sin^2(2x)) which gives cos(2x)sin(2x)/|sin(2x)|

I think it’s slightly false on the derivative

[sonic marsh]

Absolutely, but
f(x) = x+|sin(2x)|

Deriving thus results in

f'(x) = 1 + sin(x)cos(x)/|sin(x)|

Oooooooo

I see

[strong marten]

Yes but the |sin(2x)| becomes cos(2x)sin(2x)/|sin(2x)|

[sonic marsh]

Yeah, I see, i lost the 2x. Let me correct it.

I see my mistake, but thats still not the main question.

My bigger issue is how can i calculate the sign interval

Even if it is

|sin(2x)| + sin(2x)cos(2x)

[strong marten]

Yes I see

Well consider the function without the derivative, x+|sin(2x)| with x positive if 2x is on the interval [0,pi/2] mod pi it increases ( trig circle ) and if it’s on the interval [pi/2,pi] it decreases, then on the interval [pi,3pi/2] it reincreases because of the |.| and redecreases on the interval [3pi/2,2pi] etc…

Wait never mind I’m dumb

[sonic marsh]

I feel you. I have been stuck on this for the past hour lol

[strong marten]

Well if -sin(2x)cos(2x)>|sin(x)| that means -sin(2x)cos(2x)<-sin(x) or -sin(2x)cos(2x)>sin(x)

If you consider the fist case you have sin(2x)cos(2x)>sin(x) thus 2cos(x)sin(x)cos(2x)>sin(x) thus sin(x)(2cos(x)cos(2x)-1)>0

If sin(x)>0 that means x is on the interval [0,pi], but that also means 2cos(x)cos(2x)-1>0

I think you have to isolate each seperate case and study it

[sonic marsh]

I see. Give me a second while I recalcualte the derivative and solve with trig expressions. I fear I might be trying to calculate something I dont need to

[strong marten]

Also 2cos(x)cos(2x)=2cos(x)(2cos^2–1)

=4cos^3(x)-2cos(x)

There may be an easier way to do it but I can’t think of one lol

You can try studying the function 4x^3-2x-1

[sonic marsh]

Redoing seems to have simplified the issue. Still need to figure out how to get the interval.

$f'(x) = 1 + \frac{2\sin(2x)cos(2x)}{|\sin(2x)|} = 1 + \frac{\sin(4x)}{|\sin(2x)|} = \frac{|\sin(2x)| + \sin(4x)}{|\sin(2x)|} \$

[old owl]

Lonac

[strong marten]

Good luck

[sonic marsh]

Thanks for the help anyway