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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 square root of 2 minus cosine of the quantity x over 2 intersect on the i class=

Respuesta :

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

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

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

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

This is true for:

[tex]\begin{gathered} \frac{x}{2}=45\degree\Rightarrow x=90\degree \\ \frac{x}{2}=315\degree\Rightarrow x=630\degree \end{gathered}[/tex]

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

Answer: 90°

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