Question

Sin^2(x/2)=sin^2x Find the value of x

Answer 1

$\sin^2\dfrac x2=\sin^2x$
$\dfrac{1-\cos x}2=1-\cos^2x$
$1-\cos x=2-2\cos^2x$
$2\cos^2x-\cos x-1=0$
$(2\cos x+1)(\cos x-1)=0$
$\implies\begin{cases}2\cos x+1=0\\\cos x-1=0\end{cases}$$\implies\begin{cases}\cos x=-\frac12\\\cos x=1\end{cases}$

The first case occurs in $0\le x$ for $x=\dfrac{2\pi}3$ and $x=\dfrac{4\pi}3$. Extending the domain to account for all real $x$, we have this happening for $x=\dfrac{2\pi}3+2n\pi$ and $\dfrac{4\pi}3+2n\pi$, where $n\in\mathbb Z$.

The second case occurs in $0\le x$ when $x=0$, and extending to all reals we have $x=2n\pi$ for $n\in\mathbb Z$, i.e. any even multiple of $\pi$.