Trigonometry
Solve the following equation for x if x lies in the interval [−π2,π2]\left[-\frac{\pi}{2},\right.\left.\frac{\pi}{2}\right\rbrack[−2π,2π].
y=−8+3sinxy=-8+3\sin xy=−8+3sinx
x=sin−1(y+83)x=\sin^{-1}\left(\frac{y+8}{3}\right)x=sin−1(3y+8)
x=sin−1(y−83)x=\sin^{-1}\left(\frac{y-8}{3}\right)x=sin−1(3y−8)
x=sin−1(8y3)x=\sin^{-1}\left(\frac{8y}{3}\right)x=sin−1(38y)
x=sin−1(8−y3)x=\sin^{-1}\left(\frac{8-y}{3}\right)x=sin−1(38−y)