## anonymous one year ago solve between 0 and 2pi cotxsin^2x=2cotx

1. jdoe0001

$$\large { cot(x)sin^2(x)=2cot(x)\implies cot(x)sin^2(x)-2cot(x)=0 \\ \quad \\ cot(x)[sin^2(x)-2]=0\implies \begin{cases} cot(x)=0\\ sin^2(x)-2=0 \end{cases} }$$ solve those for "x", using the inverse trigonometric function, or just using your Unit Circle

2. Mertsj

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

3. jdoe0001

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