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

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 $$2\cos x - 2\cos(2x)\ge0$$ ?

• Set the derivative equal to zero and solve using the fact that $\cos{(2x)}=2\cos^2{(x)}-1$ – Peter Foreman Jun 20 at 16:49

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.

• Why should I consider p(x)? Behaviour of x is different from cos(x) in [0,2pi]. Thanks for your effort. – Cirelli94 Jun 21 at 8:22
• Because $2\cos(x)-2\cos(2x)=p\bigl(\cos(x)\bigr)$. Since $p(x)\geqslant0$ if and only if $-\frac12\leqslant x\leqslant1$, you know that $2\cos(x)-2\cos(2x)\geqslant0$ if and only if $-\frac12\leqslant\cos(x)\leqslant1$. Can you take it from here? – José Carlos Santos Jun 21 at 9:54
• but cos(x) behaves differently from x in [0,2π] ! I can't understand why you can use p(x) instead of p(cos(x)) ... cos(x) even becomes negative in that interval... – Cirelli94 Jun 24 at 8:32
• The range of $\cos$ is $[-1,1]$ and the restriction pf $p(x)$ to $[-1,1]$ is negative in $\left[-1,-\frac12\right)$ and positive on $\left(-\frac12,1\right]$. – José Carlos Santos Jun 24 at 9:22

Hint:

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

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

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?