<s>[INST] Sum of all real values of $x$ in trigonometric equation of cotangent The sum of all real values of $x$ in $\left[0,2\pi\right]$ which satisfy the equation $3\cot^2 x+8\cot x+3=0$
on solving $\displaystyle \cot x = \frac{-8\pm \sqrt{64-36}}{6} = \frac{-8\pm 2\sqrt{7}}{6} = -\frac{4}{3}\pm \frac{\sqrt{7}}{3}$
I want to go further could some help me with this, Thanks
 [/INST] Check
$$\frac{-8 \pm \sqrt{28}}{6} = \frac{-8 \pm \sqrt{4 \times 7}}{6}= \frac{-4}{3} \pm \frac{\sqrt{7}}{3}$$
"The sum of all real values of x...which satisfy the equation" haven't you found the solutions to the $\cot$ quadratic? So what is the sum of these (two) values?
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