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Where do the graphs of f of x equals cosine of the quantity x over 2 and g of x equals square root of 2 minus cosine of the quantity x over 2 intersect on the interval [0, 360°)?

Where do the graphs of f of x equals cosine of the quantity x over 2 and g of x equals-example-1
User CBRRacer
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1 Answer

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Given the functions f(x) and g(x) defined as:


\begin{gathered} f(x)=\cos ((x)/(2)) \\ g(x)=\sqrt[]{2}-\cos ((x)/(2)) \end{gathered}

The intersection of the graphs will be at f(x) = g(x):


\begin{gathered} \cos ((x)/(2))=\sqrt[]{2}-\cos ((x)/(2)) \\ 2\cos ((x)/(2))=\sqrt[]{2} \\ \cos ((x)/(2))=\frac{\sqrt[]{2}}{2}=\frac{1}{\sqrt[]{2}} \end{gathered}

This is true for:


\begin{gathered} (x)/(2)=45\degree\Rightarrow x=90\degree \\ (x)/(2)=315\degree\Rightarrow x=630\degree \end{gathered}

But only x = 90° is within the interval [0°, 360°)

Answer: 90°

User Jordi Flores
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