How can I find points of max and min of $2\sin(x)-\sin(2x)$?

Question

How can I find points of max and min of $2\sin(x)-\sin(2x)$ in $[0, 2π]$ ?

In fact the derivative is $2\cos x - 2\cos(2x)$, which I can't check with an inequality where is bigger or lesser than zero.

How can I solve $2\cos x - 2\cos(2x)\ge0$ ?

Answer 1

Note that$$2\cos(x)-2\cos(2x)=2\cos(x)-2\bigl(2\cos^2(x)-1\bigr)=-4\cos^2(x)+2\cos(x)+2.$$So, considere the polynomial function $p(x)=-4x^2+2x+2$ and check its sign.

Answer 2

Hint:

$$\cos(2x) = 2\cos^2 x - 1$$

Can you take it from here, now that you have a quadratic equation in $\cos(x)$?

Answer 3

By AM-GM $$|2\sin{x}-\sin2x|=\sqrt{4\sin^2x(1-\cos{x})^2}=2\sqrt{(1-\cos{x})^3(1+\cos{x})}=$$ $$=6\sqrt3\sqrt{\left(\frac{1-\cos{x}}{3}\right)^3(1+\cos{x})}\leq6\sqrt3\sqrt{\left(\frac{3\cdot\frac{1-\cos{x}}{3}+1+\cos{x}}{4}\right)^4}=\frac{3\sqrt3}{2}.$$ The equality occurs for $\frac{1-\cos{x}}{3}=1+\cos{x}.$

Can you end it now?