PT $\Leftrightarrow 5[\dfrac{3}{5}\sin(x+1)+\dfrac{4}{5}\cos(x+1)]=5$ (*)
Đặt $\cos \alpha=\dfrac{3}{5};\sin\alpha=\dfrac{4}{5}$
(*) $\Leftrightarrow \sin(x+1+\alpha)=1$
$\Leftrightarrow x+1+\alpha=\dfrac{\pi}{2}+k2\pi$
$\Leftrightarrow x=-1-\alpha+\dfrac{\pi}{2}+k2\pi$