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The sum of all real values of $\theta$ in $3\cot^2 \theta+8\cot \theta+3 = 0,$ Where $0\leq \theta\leq 2\pi$

$\bf{My\; Try::}$ Solving above equation, we get $$\cot \theta = \frac{-8\pm \sqrt{28}}{6} = \frac{-8\pm 7\sqrt{2}}{6}$$

Now how can i solve it after that, Help required, Thanks

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If $\cot\theta_1,\cot\theta_2$ are the two roots,

we have $\cot\theta_1\cdot\cot\theta_2=\dfrac33$

$\iff\tan\theta_1=\cot\theta_2=\tan(90^\circ-\theta_2)$

$\implies \theta_1+\theta_2=180^\circ m+90^\circ$ where $m$ is any integer

Now as both roots are negative, the roots will lie in the second or in the fourth quadrant.

For the second quadrant, $90^\circ<\theta_1,\theta_2<180^\circ,$

$$\theta_1+\theta_2=180^\circ+90^\circ$$

For the fourth quadrant, $270^\circ<\theta_1,\theta_2<360^\circ,$

$$\theta_1+\theta_2=180^\circ\cdot3+90^\circ$$

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