TrigonometryHardTrigonometric Equations ShareTrigonometric Equations · Hard problemSolve sin2x=3cosx\sin 2x=\sqrt{3}\cos xsin2x=3cosx on [0,2π)[0,2\pi)[0,2π). (Hint: sin2x=2sinxcosx\sin 2x=2\sin x\cos xsin2x=2sinxcosx.)Check Show answer Walk through it step by step Practice more like this Review this concept