Question

At which values in the interval [0, 2π) will the functions f (x) = cos 2x + 2 and g(x) = cos x + 1 intersect?

Answer 1

`x=(\pi)/(3),(\pi)/(2),(3\pi)/(2),(5\pi)/(3)`

Explanations:

A solution lies where two functions intersect. Given the following functions:

`\begin{gathered} f(x)=\cos 2x+2 \\ g(x)=\cos (x)+1 \end{gathered}`

The function intersects at the point where f(x) = g(x) that is:

`\begin{gathered} \cos 2x+2=\cos (x)+1 \\ \cos 2x-\cos x+2-1=0 \\ \cos 2x-\cos x+1=0 \end{gathered}`

Recall from trigonometry identity that:

`\begin{gathered} \cos 2x=cos(x+x)=\cos ^2x-\sin ^2x \\ \cos 2x=\cos ^2x-(1-\cos ^2x) \\ \cos 2x=2\text{cos}^2x-1 \end{gathered}`

Substitute the expression for cos2x into the quadratic function to have:

`\begin{gathered} \cos 2x-\cos x+1=0 \\ (2\cos ^2x-1)-\text{cosx}+1=0 \\ 2\text{cos}^2x-\cos x-1+1=0 \\ 2\text{cos}^2x-\cos x=0 \end{gathered}`

Simplify the result for the value(s) of "x"

`\begin{gathered} 2\cos ^2x-\cos x=0 \\ \cos x(2\cos x-1)=0 \\ 2\cos x=1\text{ and cos x = 0} \\ \cos x=(1)/(2) \\ x=\cos ^(-1)((1)/(2)) \\ x_1=60^0=(\pi)/(3) \end{gathered}`

Since cosine is positive in the first and fourth quadrant, the other angles will be:

`\begin{gathered} x_2=360-x_1 \\ x_2=2\pi-(\pi)/(3) \\ x_2=(5\pi)/(3)^{} \end{gathered}`

For cos x = 0;

`\begin{gathered} \cos x=0 \\ x=\text{cos}^(-1)0 \\ x_3=(\pi)/(2) \end{gathered}`

Since cos is also positive in the fourth quadrant, hence;

`\begin{gathered} x_4=2\pi-(\pi)/(2) \\ x_4=(3\pi)/(2) \end{gathered}`

Therefore the values in the interval [0, 2π) that will make the functions

f (x) = cos 2x + 2 and g(x) = cos x + 1 intersect are:

`x=(\pi)/(3),(\pi)/(2),(3\pi)/(2),(5\pi)/(3)`