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? 
 A: 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 
$$ \sin \left( \frac{ \pi ( 1 - 2x + \sqrt{x} ) }{4} \right) \sin \left( \frac{ \pi ( 1 - 2x - \sqrt{x} ) }{4} \right) < 0 $$
A: Hint:
The difference between the two members is a doubly oscillating function that stays in range $[-2,2]$. The oscillations due to the RHS get slower and slower.
There is no $x$ for which the two members both equal $1$ or $-1$ (by the transcendence of $\pi$), so that all the roots are simple.
Hence, the inequality is true for all $x$ between every other pair of roots, starting from the first pair (for $x>0$).
From the work of @ILoveMath, you can see that by means of the sum-to-product formula, you can compute those roots by solving quadratic equations in $\sqrt x$.
