Finding number of solution trigonometry (1 Viewer)

BlueGas · Oct 9, 2015

Why do I have to do 3cosx? Why can't I just graph cosx and then find the solutions from there?

InteGrand · Oct 9, 2015

BlueGas said:
Why do I have to do 3cosx? Why can't I just graph cosx and then find the solutions from there?

$Because $\sin 4x =\cos x$ is equivalent to $3\sin 4x = 3\cos x$. Since you have a graph of the LHS of the latter equation, you should sketch the \textsl{RHS} of the latter equation onto your current graph to get the required number of solutions (intersection points). You don't have the graph of the LHS of the former equation, so no point sketching the RHS of that equation.$

Finding number of solution trigonometry (1 Viewer)

BlueGas

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InteGrand

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