Question

Solve for x where 0 Equation: cot^2x+2cotx+1=0

Answer 1

`\bf \stackrel{\stackrel{notice~~ax^2+bx+c=0}{\downarrow }}{\underset{\textit{tis just a quadratic}}{cot^2(x)+2cot(x)+1=0}}\implies [cot(x)+1][cot(x)+1]=0 \\\\\\ cot(x)+1=0\implies cot(x)=-1\implies cot^(-1)[cot(x)]=cot^(-1)(-1) \\\\\\ x=cot^(-1)(-1)\implies x= \begin{cases} \stackrel{II~Quadrant}{(3\pi )/(4)}\\[1em] \stackrel{IV~Quadrant}{(7\pi )/(4)} \end{cases}`