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I tried:

$$2\sin(2x)\cos^2(x) = \\ 2\cdot\cos x\cdot\sqrt{\sin x(1-\sin x)(1+\sin x)} = \\ 2\cdot \sqrt{1-\sin^2x}\cdot\sqrt{\sin x (1-\sin x)(1+\sin x)} = \\ 2 \cdot \sqrt{(1-\sin x)(1+\sin x)\sin x (1-\sin x)(1+\sin x)} = \\ 2 \cdot \sqrt{(1-\sin^2x)^2\sin x} = \\ 2 \cdot (1-\sin^2x)\sqrt{\sin x} = \\???$$

What do I do next?

Mark Read
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  • Does this help https://math.stackexchange.com/questions/164221/period-of-the-sum-product-of-two-functions? – Moo May 06 '17 at 21:06
  • You cannot use square roots. What if you choose $x=-\pi/2$? Plus if you look at your last line, and the first line, you get $\sin(2x)=\sqrt{\sin x}$ – Andrei May 06 '17 at 21:40

2 Answers2

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$$2\sin(2x)\cos^2 x=2\sin(2x)(\cos(2x)+1)/2=\sin(2x)+\frac{1}{2}\sin(4x)$$ The first term has a period of $\pi$, the second $\pi/2$, so the period for your expression is $\pi$

Andrei
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Since a period of $f(x)=2\sin 2x\cos^2 x$ is certainly $2\pi$ and $f(x)$ vanishes at $0$, $\pi/2$, $\pi$, $3\pi/2$ and $2\pi$, in the interval $[0,2\pi]$, the minimum period can only be $\pi/2$ or $\pi$, if it is not $2\pi$. Clearly $$ f(x+\pi)=2\sin(2x+2\pi)\cos^2(x+\pi)=2\sin2x(-\cos x)^2=f(x) $$

So a period is $\pi$. Can $\pi/2$ be a period? Hint: look at the sign of $f$.

egreg
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