Question

Giải pt :

\(tan.\left(x+\dfrac{\pi}{4}\right)+cot\cdot\left(2x-\dfrac{\pi}{3}\right)=0\)

Answer 1

\(tan\cdot\left(x+\dfrac{\pi}{4}\right)+cot\cdot\left(2x-\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=-cot\cdot\left(2x-\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=cot\cdot\left(-2x+\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=tan\cdot\left(\dfrac{\pi}{2}+2x-\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=tan\cdot\left(\dfrac{\pi}{6}+2x\right)\)

\(\Leftrightarrow x+\dfrac{\pi}{4}=\dfrac{\pi}{6}+2x+k\pi\)

\(\Leftrightarrow-x=\dfrac{-\pi}{12}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{12}-k\pi\left(k\in Z\right)\)