Fórmulas a usar (producto a suma):
$$ \cos A\,\sen B=\tfrac12\big[\sen(B{+}A)+\sen(B{-}A)\big]. $$

Aplicando a ambos lados:
$$ \begin{aligned} \cos x\,\sen 7x &=\tfrac12\big[\sen(7x+x)+\sen(7x-x)\big] =\tfrac12\big[\sen 8x+\sen 6x\big],\\ \cos 3x\,\sen 5x &=\tfrac12\big[\sen(5x+3x)+\sen(5x-3x)\big] =\tfrac12\big[\sen 8x+\sen 2x\big]. \end{aligned} $$
Igualando:
$$ \sen 8x+\sen 6x=\sen 8x+\sen 2x \quad\Longrightarrow\quad \sen 6x=\sen 2x. $$

Resolver \(\sen 6x=\sen 2x\):
$$ \sen 6x-\sen 2x=2\cos\!\left(\tfrac{6x+2x}{2}\right)\sen\!\left(\tfrac{6x-2x}{2}\right) =2\cos(4x)\,\sen(2x)=0. $$
Por tanto,
$$ \boxed{\ \sen(2x)=0\ } \quad \text{o} \quad \boxed{\ \cos(4x)=0\ }. $$
De cada caso:
$$ \sen(2x)=0 \Rightarrow 2x=n\pi \Rightarrow \boxed{x=\dfrac{n\pi}{2}},\qquad \cos(4x)=0 \Rightarrow 4x=\dfrac{\pi}{2}+n\pi \Rightarrow \boxed{x=\dfrac{\pi}{8}(2n+1)}. $$

Respuesta:
$$ \boxed{\,x=\dfrac{n\pi}{2}\ \text{o}\ x=\dfrac{\pi}{8}(2n+1),\quad n\in\mathbb{Z}. \,} $$