205k views
3 votes
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 Jed Schaaf
by
8.4k points

1 Answer

5 votes

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 Colla
by
8.1k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories