Mathematics - Trigonometric Equation Question with Solution | TestHub

Let $u=\cot x$. Then the equation becomes $3u^2+8u+3=0$.

Both roots are real, and their product is $1$.

But $\cot x$ is a bijection in $(0,\pi )$. Let $x_1,x_2$ be roots such that $0<x_1,x_2<\pi$. Since $\cot x_1$ and $\cot x_2$ are both negative,

$\frac{\pi }{2}<x_1,x_2<\pi$.

But $\pi <x_1+x_2<2\pi$.

$\cot x_1\cot x_2=1$,

$\cot x_1\cdot \cot \left(\frac{3\pi }{2}-x_1\right)=1$.

$x_1+x_2=\frac{3\pi }{2}$. Similarly, $x_3+x_4=\frac{7\pi }{2}$.

$k=5\pi$.