<s>[INST] Trigonometric inequality $\sin(\pi x)>\cos(\pi \sqrt x)$ Solve the inequality 
$\sin(\pi x)>\cos(\pi \sqrt x)$ 
I don't know where to begin. Hints? 
 [/INST] Thought:
Write $\sin (\pi x ) - \cos( \pi \sqrt{x} ) > 0$. Note $\sin ( \pi x) = \cos ( \pi/2 - \pi x) = \cos \left( \frac{ \pi - 2 \pi x }{2} \right) $. Thus,
$$ \cos \left( \frac{ \pi - 2 \pi x }{2} \right) - \cos( \pi \sqrt{x} ) = - 2 \sin \left( \frac{ \frac{\pi - 2 \pi x}{2} + \pi \sqrt{x}  }{2} \right)\sin \left( \frac{ \frac{\pi - 2 \pi x}{2} - \pi \sqrt{x}  }{2} \right) =$$
$$ = - 2 \sin \left( \frac{ \pi ( 1 - 2x + \sqrt{x} ) }{4} \right) \sin \left( \frac{ \pi ( 1 - 2x - \sqrt{x} ) }{4} \right) $$
And this is at least $0$ iff 
$$ \sin \left( \frac{ \pi ( 1 - 2x + \sqrt{x} ) }{4} \right) \sin \left( \frac{ \pi ( 1 - 2x - \sqrt{x} ) }{4} \right) < 0 $$
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