Trig and DE....

Using $\ sin^2 x +cos^2 x = 1$ : $\frac{\sqrt{2+2sinx - cos^2 x} }{cosx siny} = \frac{\sqrt{sin^2 + 2sinx +1}}{cosx siny} = \frac{sinx +1}{cosx siny} = \frac{dy}{dx}$ This differential equation is separable such that: $\int siny \ dy = \int tanx +secx \ dx$ Now the constant of integration is zero so we have: $-cos y = Ln|secx| + Ln|secx +tanx| = Ln|sec^2 x +secx tanx|$ $= Ln| \frac{sinx +1}{cos^2 x}| = Ln| \frac{sinx+1}{1-sin^2 x }| = Ln| \frac{1}{1-sinx} | = -Ln|1-sinx|$ $\Rightarrow\ cosy = Ln|1-sinx|$ Now we substitute y = 75 degrees into cosy: $e^{cos75} = 1-sinx \therefore\ x = sin^{-1} (1-e^{cos75}) = 197 \ (3sf)$