Question

Giải pt:

Answer 1

ĐKXĐ: \(\frac{2x+1}{6}<>k\pi\)

=>\(2x+1<>6k\pi\)

=>\(2x<>6k\pi-1\)

=>\(x<>3k\pi-\frac12\)

\(\cot\left(\frac{2x+1}{6}\right)=\tan\left(\frac12\right)\)

=>\(\cot\left(\frac{2x+1}{6}\right)=\cot\left(\frac{\pi}{2}-\frac12\right)\)

=>\(\frac{2x+1}{6}=\frac{\pi-1}{2}+k\pi\)

=>\(2x+1=3\left(\pi-1\right)+6k\pi\)

=>\(2x=3\pi-3-1+6k\pi=3\pi+6k\pi-4\)

=>\(x=\frac32\pi+3k\pi-2\) (nhận)