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

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

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  • $\begingroup$ Why should I consider p(x)? Behaviour of x is different from cos(x) in [0,2pi]. Thanks for your effort. $\endgroup$ – Cirelli94 Jun 21 at 8:22
  • $\begingroup$ 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? $\endgroup$ – José Carlos Santos Jun 21 at 9:54
  • $\begingroup$ 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... $\endgroup$ – Cirelli94 Jun 24 at 8:32
  • $\begingroup$ 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 $\left(-\frac12,1\right]$. $\endgroup$ – José Carlos Santos Jun 24 at 9:22
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Hint:

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

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

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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?

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