Question

solve between 0 and 2pi
cotxsin^2x=2cotx

Answer 1

\(\large {
cot(x)sin^2(x)=2cot(x)\implies cot(x)sin^2(x)-2cot(x)=0
\\ \quad \\
cot(x)[sin^2(x)-2]=0\implies
\begin{cases}
cot(x)=0\\
sin^2(x)-2=0
\end{cases}
}\)

solve those for "x", using the inverse trigonometric function, or just using your Unit Circle

Answer 2

1. Subtract 2cotx from both sides
2. Factor out cotx
3. Set each factor equal to 0 and solve each equation.

Answer 3

btw, if you didn't get the lines from surjithayer on the previous posting
you can just let us know that you dunno, or what part you don't get, and maybe just explain you'd repost to get more eyes