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?
$\endgroup$ 22 Answers
$\begingroup$$$2\sin(2x)\cos^2 x=2\sin(2x)(\cos(2x)+1)/2=\sin(2x)+\frac{1}{2}\sin(4x)$$ The first term has a period of $\pi$, the second $\pi/2$, so the period for your expression is $\pi$
$\endgroup$ 3 $\begingroup$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 $f$.
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