Find the maximum and minimum values of $x-\sin2x+\frac{1}{3}\sin 3x$ in $[-\pi,\pi]$

Question

Find the maximum and minimum values of $x-\sin2x+\frac{1}{3}\sin 3x$ in $[-\pi,\pi]$.

---

Let $f(x)=x-\sin2x+\frac{1}{3}\sin 3x$

$f'(x)=1-2\cos2x+\cos3x$

Put $f'(x)=0$

$1-2\cos2x+\cos3x=0$ gives $2\sin^2x-\cos2x+\cos3x=0$

I am stuck here.I cannot find the critical values and the maximum and minimum values of the function.Please help me.

Answer 1

Hint:

$$f'(x)=1-2(2\cos^2x-1)+4\cos^3 x-3\cos x$$ $$=4\cos^3 x-4\cos^2 x-3\cos x+3$$

For $f'(x)=0$, $\cos x=1$ is a solution.

Answer 2

Continuing from $1-2\cos 2x+\cos 3x=0$:

$1-2\cos 2x+\cos 3x=0$

$1-2\left(2\cos ^2x-1\right)+\cos \left(x+2x\right)=0$

$1-2\left(2\cos ^2x-1\right)+\cos x\cos 2x-\sin x\sin 2x=0$

$1-2\left(2\cos ^2x-1\right)+\cos x\left(2\cos ^2x-1\right)-\sin x\left(2\sin x\cos x\right)=0$

$1-2\left(2\cos ^2x-1\right)+\cos x\left(2\cos ^2x-1\right)-2\sin ^2x\cos x=0$

$1-2\left(2\cos ^2x-1\right)+\cos x\left(2\cos ^2x-1\right)-2\left(1-\cos ^2x\right)\cos x=0$

$1-4\cos ^2x+2+2\cos ^3x-\cos x-2\cos x+2\cos ^3x=0$

$4\cos ^3x-4\cos ^2x-3\cos x+3=0$

$4\cos ^2x\left(\cos x-1\right)-3\left(\cos x-1\right)=0$

$\left(4\cos ^2x-3\right)\left(\cos x-1\right)=0$

$\left(\cos ^2x-\frac34\right)\left(\cos x-1\right)=0$

$\left(\cos x-\frac{\sqrt3}2\right)\left(\cos x+\frac{\sqrt3}2\right)\left(\cos x-1\right)=0$

$\cos x=\frac{\sqrt3}2$ or $-\frac{\sqrt3}2$ or $1$.