điều kiện xác định \(\left\{{}\begin{matrix}cos\left(3x-\dfrac{\pi}{2}\right)\ne0\\sinx\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x-\dfrac{\pi}{2}\ne\dfrac{\pi}{2}+k2\pi\\x\ne k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{3}+\dfrac{2}{3}k\pi\\x\ne k\pi\end{matrix}\right.\) \(\left(k\in Z\right)\)
ta có : \(tan\left(3x-\dfrac{\pi}{2}\right)+cotx=0\)
\(\Leftrightarrow tan\left(3x-\dfrac{\pi}{2}\right)+cot\left(\dfrac{\pi}{2}-\left(\dfrac{\pi}{2}-x\right)\right)=0\)
\(\Leftrightarrow tan\left(3x-\dfrac{\pi}{2}\right)-tan\left(\dfrac{\pi}{2}-x\right)=0\)
\(\Leftrightarrow tan\left(3x-\dfrac{\pi}{2}\right)=tan\left(\dfrac{\pi}{2}-x\right)\) \(\Leftrightarrow3x-\dfrac{\pi}{2}=\dfrac{\pi}{2}-x+k\pi\Leftrightarrow4x=\pi+k\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{4}\left(k\in Z\right)\left(tmđk\right)\)
vậy phương trình có một hệ nghiệm duy nhất là \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{4}\)
