Hi, Can someone please show me how to answer this question: Solve 2sin^2x = sin2x, 0<x<pi Thanks! p.s I know its simple, but I'm stuck for some reason..
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Originally Posted by CSG18 Hi, Can someone please show me how to answer this question: Solve 2sin^2x = sin2x, 0<x<pi Thanks! p.s I know its simple, but I'm stuck for some reason.. Hi CSG18, r\:\: \sin x - \cos x=0" alt="2\sin x=0 \:\r\:\: \sin x - \cos x=0" /> r\:\: \sin x = \cos x" alt="\sin x=0 \:\r\:\: \sin x = \cos x" /> r \:\:\boxed{x=\frac{\pi}{4}}" alt="\boxed{x=0} \:\r \:\:\boxed{x=\frac{\pi}{4}}" />
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