Mathematics - Trigonometric Equation Question with Solution | TestHub

${\sin }^2\left(\frac{x}{2}\right)=\frac{x}{(2n+1)\pi }$

the graph of ${\sin }^2\left(\frac{x}{2}\right)$ will be above the $x$-axis and will be meeting the $x$-axis at $0,2\pi ,4\pi ,\dots$ etc. It will attain maximum values at odd multiples of $\pi$ lie. $\pi ,3\pi ,\dots (2n+1)\pi$. The last point after which graph of $\mathrm{y}=\frac{\mathrm{x}}{(2\mathrm{n}+1)\pi }$ will stop cutting will be $(2\mathrm{n}+1)\pi$.

Total intersection $=2(\mathrm{n}+1)$