$12\sin2x+5\cos2x=13$
$\Leftrightarrow \dfrac{12}{13}\sin2x+\dfrac{5}{13}\cos2x=1$
Đặt $\cos\alpha=\dfrac{12}{13};\sin\alpha=\dfrac{5}{13}$
$\Rightarrow \sin(2x+\alpha)=1$
$\Leftrightarrow 2x+\alpha=\dfrac{\pi}{2}+k2\pi$
$\Leftrightarrow x=\dfrac{-\alpha}{2}+\dfrac{\pi}{4}+k\pi$