Let [tex]f(x) = |\cos x|+|\cos 2x|[/tex]. Observe that [tex]f(x)=f(x+2\pi)[/tex], so we need only consider [tex]x\in [0,2\pi)[/tex]. Also [tex]f(x)=f(2\pi-x)[/tex] so it is enough to consider [tex]x\in[0,\pi][/tex]. Finally [tex]f(x)=f(\pi-x)[/tex] so we only need to consider [tex]x\in[0,\pi/2][/tex].
When [tex]x\in[0,\pi/4][/tex] we have [tex]f(x)=\cos x+\cos 2x[/tex]. Clearly as [tex]x[/tex] increases [tex]f(x)[/tex] decreases so the minimum occurs at [tex]f(\pi/4) = \sqrt{2}/2[/tex].
When [tex]x\in[\pi/4,\pi/2][/tex] we have [tex]f(x)=\cos x-\cos 2x = -2y^2+y+1[/tex] where [tex]y = \cos x[/tex]. Clearly the minimum value occurs when [tex]y[/tex] is as large or as small as possible, i.e. [tex]x=\pi/4[/tex] or [tex]\pi/2[/tex]. It is easy to check the minimum is at [tex]x=\pi/4[/tex], consequently [tex]f(x)\geq f(\pi/4) = \sqrt{2}/2[/tex] for all [tex]x[/tex].
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