Dr. Fredrick, I misread an identity, sorry.
Here is the correct problem and answers.
\(\displaystyle tan(2x)-cot(x) \ = \ 0\)
\(\displaystyle \frac{sin(2x)}{cos(2x)}-\frac{cos(x)}{sin(x)} \ = \ 0\)
\(\displaystyle \frac{2sin(x)cos(x)}{2cos^{2}(x)-1}-\frac{cos(x)}{sin(x)} \ = \ 0\)
\(\displaystyle \frac{2sin^{2}(x)cos(x)-cos(x)[2cos^{2}(x)-1]}{[2cos^{2}(x)-1]sin(x)]} \ = \ 0\)
\(\displaystyle \frac{2[1-cos^{2}(x)]cos(x)-2cos^{3}(x)+cos(x)}{[2cos^{2}(x)-1]sin(x)} \ = \ 0\)
\(\displaystyle \frac{2cos(x)-2cos^{3}(x)-2cos^{3}(x)+cos(x)}{[2cos^{2}(x)-1]sin(x)} \ = \ 0\)
\(\displaystyle \frac{3cos(x)-4cos^{3}(x)}{[2cos^{2}(x)-1]sin(x)} \ = \ 0\)
\(\displaystyle \frac{cos(x)[3-4cos^{2}(x)]}{[2cos^{2}(x)-1]sin(x)} \ = \ 0\)
\(\displaystyle cos(x) \ = \ 0, \ \implies \ x \ = \ \frac{\pi}{2}, \ \frac{3 \pi}{2}, \ cos^{2}(x) \ = \ \frac{3}{4}, \ \implies \ x \ = \ \frac{\pi}{6}, \ \frac{5 \pi}{6}, \ \frac{7 \pi}{6}, \ \frac{11 \pi}{6}\)
\(\displaystyle Hence, \ all \ are \ solutions, \ again \ I'm \ sorry.\)
