TrigonometryHardTrigonometric Equations ShareTrigonometric Equations · Hard problemSolve 2sin2x−(2+2)sinx+2=02\sin^2 x-(2+\sqrt{2})\sin x+\sqrt{2}=02sin2x−(2+2)sinx+2=0 on [0,2π)[0,2\pi)[0,2π).Check Show answer Walk through it step by step Practice more like this Review this concept