Find all extrema of a complicated trigonometric function

Question

Problem

Find all local extrema for

$$f(x) = \frac{\sin{3x}}{1+\frac{1}{2}\cos{3x}}$$

Attempted solution

My basic approach is to take the derivative, set the derivative equal to zero and solve for x.

Taking the derivative with the quotient rule and a few cases of the chain rule for the trigonometric functions with a final application of the Pythagorean identity:

$$f'(x) = \frac{(1+\frac{1}{2}\cos{3x})(3\cos{3x})+1.5\sin{3x}\sin {3x}}{(1+\frac{1}{2}\cos{3x})^2} = \frac{3\cos 3x+1.5\cos^2 3x + 1.5\sin^2 3x}{(1+\frac{1}{2}\cos 3x)^2} = \frac{3 \cos 3x + 1}{(1+\frac{1}{2}\cos 3x)^2}$$

Putting it equal to zero and solving for x:

$$3\cos 3x + 1 = 0 \Rightarrow x = \frac{\arccos{\Big(-\frac{1}{3}\Big)}}{3} = \frac{\pi}{6} + \frac{2\pi n}{3}$$

...however the expected answer is $\pm\frac{2\pi}{9} + \frac{2\pi n}{3}$

So I must have gone wrong somewhere.

Answer 1

It suffices to cancel the numerator of the derivative,

$$\cos(3x)(2+\cos(3x))+\sin(3x)\sin(3x)=2\cos(3x)+1=0$$

and

$$3x=2k\pi\pm\frac{2\pi}3.$$

Answer 2

It should be $$f'(x) = \frac{(1+\frac{1}{2}\cos{3x})(3\cos{3x})-\frac{3}{2}\sin{3x}\sin {3x}}{(1+\frac{1}{2}\cos{3x})^2}.$$

I like the following way.

Let $x=\frac{2\pi}{9}.$

Thus, we get a value $\frac{2}{\sqrt3}.$

We'll prove that it's a maximal value.

Indeed, we need to prove that $$\frac{\sin3x}{2+\cos3x}\leq\frac{1}{\sqrt3}$$ or $$\sqrt3\sin3x-\cos3x\leq2,$$ which is true by C-S: $$\sqrt3\sin3x-\cos3x\leq\sqrt{((\sqrt3)^2+(-1)^2)(\sin^23x+\cos^23x)}=2.$$ By the same way we can get a minimal value.

I got $-\frac{2}{\sqrt3},$ which occurs for $x=-\frac{2\pi}{9}.$

Answer 3

Looks like you have a problem with the differentiation. You should have.

You dropped a factor of 3 in the right-hand term. It should be $(\frac 32 \sin 3x)(\sin 3x)$ in the first line.

You have brought it back by the time you get to.

$3\cos 3x + 1.5\cos^2 3x + \frac 12 \sin^2 3x = 0$

But then $1.5$ becomes $1$ in the next line.

$3\cos 3x + 1.5 = 0$

Solving for x:

$\cos3x = -\frac 12$

$3x = \pm\frac {2\pi}{3} + 2n\pi\\ x = \pm \frac {2\pi}{9} + \frac {2n\pi}{3}$

$x = \frac {2\pi}{9} + \frac {2n\pi}{3}$ are the maxima and $x = -\frac {2\pi}{9} + \frac {2n\pi}{3}$ are the minima

Answer 4

This function has period $\frac{2\pi}3$ and it is an odd function, so we need to determine its variations only on $\bigl[0,\frac\pi 3\bigr]$.

Now simplifying the derivative, you get $$f'(x)=\frac{3\bigl(\frac12+\cos 3x\bigr)}{\bigl(1+\frac{1}{2}\cos{3x}\bigr)^2},$$ which has the sign of $\;\frac12+\cos 3x$, so we have to solve the inequation $$\cos 3x>-\tfrac 12\quad\text{on}\quad \bigl[0,\tfrac\pi 3\bigr].$$ Can you continue?