Relative extremes using periodicity

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If $f$ is a periodic function, then the location of all absolute extrema in the interval $(−\infty, +\infty)$ can be obtained by finding the absolute extremes in a period and using the radiodicity to locate the others. Use this idea in these exercises to find the values absolute extrema of the function and indicate the values ​​of $x$ at which they occur.

$(a) \, \sin x$.
$(b) \, 2\cos x + \cos 2x$.

For letter $(a)$ I thought about locating the absolute extremes by doing $(\sin(x))' = \cos(x) = 0$, that is, $$x=\frac{\pi}{2}$$ or $$x =\frac{3 \pi}{2}$$

Making $(\cos(x))'=-\sin(x)$, and then $$\sin(\pi/2)=-1$$ That is, a relative maximum. And $$\sin(3\pi/2)=1$$ In other words, a relative minimum. Using periodicity for the others, we have that every point in the form $$\frac{\pi+2k}{2}$$ is a relative maximum and every point in the form $$\frac{3\pi+2k}{2 }$$ is a relative minimum.

Trying to apply this same reasoning to letter $(b)$, I have the first derivative as $$2(-\sin(x))+(-\sin(2x))2$$ But I can't find the roots of this equation. How should I proceed in this case?

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Comment for $a$, you have to define what is $k$.

For part $b$, $$-2\sin(x) -2\sin(2x)=0$$

$$\sin(x) + 2\sin(x)\cos(x)=0$$

Hence we have either $\sin(x)=0$ or $1+2\cos(x)=0$.

Can you take it from here?