Вычислите sinx если sin x/2 = 3/5 0

Решите задачу:

$sinx - ? , sin \frac{x}{2} = \frac{3}{5} \\ sin^{2}\frac{x}{2} = \frac{1-cosx}{2}= \frac{9}{25} \\ \frac{1-cosx}{2}= \frac{9}{25} | *50 \\ 25 - 25cosx = 18 \\ -25cosx = -7 \\ cosx = \frac{7}{25} ; \\ =\ \textgreater \ sin^{2}x + cos^{2}x = 1 \\ sin^{2}x = 1 - cos^{2}x \\ sin^{2}x = 1 - \frac{49}{625} \\ sin^{2}x = \frac{576}{625} \\ =\ \textgreater \ sinx = \frac{24}{25} \\ or (ili) \\ sinx = - \frac{24}{25} \\ \\ x_{1} = (-1)^{n} * arcsin \frac{24}{25} + \pi n \\ x_{2} = (-1)^{n+1} * arcsin \frac{24}{25} + \pi n \\$