Question

If $sinx=\frac{\sqrt{5}}{3}$ and $\frac{\pi }{2}<x<\pi$, find the values of $sinx=2sin\frac{x}{2}cos\frac{x}{2}$

Answer 1

$1)sinx=2sin\frac{x}{2}cos\frac{x}{2}\frac{\sqrt{5}}{3}=2si{n}^2\frac{x}{2}\sqrt{1-si{n}^2\frac{x}{2}}\frac{5}{9}=4si{n}^2\frac{x}{2}(1-si{n}^2\frac{x}{2})\frac{5}{9}=4si{n}^2\frac{x}{2}-4si{n}^2\frac{x}{2}4si{n}^4\frac{x}{2}-4si{n}^2\frac{x}{2}+\frac{5}{9}=036si{n}^4\frac{x}{2}-36si{n}^2\frac{x}{2}+5=0$
$si{n}^2\frac{x}{2}=\frac{36\pm \sqrt{{36}^2-4\left(36\right)\left(5\right)}}{72}=\frac{36\pm \sqrt{576}}{72}si{n}^2\frac{x}{2}=\frac{36\pm 24}{72}si{n}^2\frac{x}{2}=\frac{60}{72}si{n}^2\frac{x}{2}=\frac{12}{72}=\frac{1}{6}si{n}^2\frac{x}{2}=\frac{5}{6}sin\frac{x}{2}=\sqrt{\frac{1}{6}}\Rightarrow sin\frac{x}{2}=\sqrt{\frac{5}{6}}$