Mathematics - Trigonometric Equation Question with Solution | TestHub

$2{\cot }^{2}x-5\mathrm{cosec}x=1$

$$2{\mathrm{cosec}}^{2}x-5\mathrm{cosec}x-3=0$$

$$2(\mathrm{cosec}x+1)(\mathrm{cosec}x-3)=0$$

$$\mathrm{cosec}x=-\frac{1}{2},\mathrm{cosec}x=3$$

$\mathrm{cosec}x=3$ gives solutions in the $1^{st}$ and $2^{nd}$ quadrants, while $\mathrm{cosec}x=-\frac{1}{2}$ gives no solution. So, in $[0,2\pi ]$, we get only two solutions. In $[0,6\pi ]$, we get 6 solutions, and between $6\pi$ and $7\pi$, we get the seventh solution. Hence, $n=7$.