[tex]\sin x\ge0[/tex] for all [tex]x\in[0,\pi][/tex], while [tex]\cos x>0[/tex] for [tex]0<x<\dfrac\pi2[/tex] and [tex]\cos x<0[/tex] for [tex]\dfrac\pi2<x<\pi[/tex]. So you already know that [tex]4\sin x>10\cos x[/tex] over the second half of the interval.
In the first half, [tex]4\sin x[/tex] is an increasing function, from [tex]4\sin0=0[/tex] to [tex]4\sin\dfrac\pi2=4[/tex]. Meanwhile [tex]10\cos x[/tex] is a decreasing function, from [tex]10\cos0=10[/tex] to [tex]10\cos\dfrac\pi2=0[/tex]. Therefore there must be a point between 0 and [tex]\dfrac\pi2[/tex] where the two curves intersect. So we find it:
[tex]4\sin x=10\cos x\implies \tan x=\dfrac{10}4=\dfrac52\implies x=\arctan\dfrac52\approx1.1903[/tex]
So we know that [tex]10\cos x>4\sin x[/tex] in [tex]\left[0,\arctan\dfrac52\right)[/tex], and [tex]4\sin x>10\cos x[/tex] in [tex]\left(\arctan\dfrac52,\pi\right][/tex].
The area between the curves is then
[tex]\displaystyle\int_0^{\arctan(5/2)}(10\cos x-4\sin x)\,\mathrm dx+\int_{\arctan(5/2)}^\pi(4\sin x-10\cos x)\,\mathrm dx=4\sqrt{29}[/tex]
