Question

3arcsin^2x+5π arcsinx-2π^2=0​

Answer 1

$3 \arcsin {}^{2} (x) + 5\pi \arcsin(x) - 2 {\pi}^{2} = 0 \\ \arcsin(x) = t , \: t \in [-1;1]\\ 3 {t}^{2} + 5\pi t - 2 {\pi}^{2} = 0 \\ D = (5\pi) {}^{2} - 4 \times 3( - 2\pi {}^{2} ) = 25 {\pi}^{2} + 24 {\pi}^{2} = 49{\pi}^{2} \\ t_{1} = \frac{ - 5\pi + 7\pi}{6} = \frac{\pi}{3} \\ t_{2} = \frac{ - 5\pi - 7\pi}{6} = - 2\pi \\ t_{2} \notin[-1;1]\Rightarrow t = \frac{\pi}{3} \\ \arcsin(x) = \frac{\pi}{3} \\ x = \frac{ \sqrt{3} }{2}$