Question

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

Answer 1

`\sin^2\frac x2=\sin^2x`

`\frac{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<2\pi` for
`x=\frac{2\pi}3` and
`x=\frac{4\pi}3`. Extending the domain to account for all real
`x`, we have this happening for
`x=\frac{2\pi}3+2n\pi` and
`\frac{4\pi}3+2n\pi`, where
`n\in\mathbb Z`.

The second case occurs in
`0\le x<2\pi` 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`.