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

Answer:

$x=\frac{\pi}{3},\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{3}$

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"

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

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

$\begin{gathered} x_2=360-x_1 \\ x_2=2\pi-\frac{\pi}{3} \\ x_2=\frac{5\pi}{3}^{} \end{gathered}$

For cos x = 0;

$\begin{gathered} \cos x=0 \\ x=\text{cos}^{-1}0 \\ x_3=\frac{\pi}{2} \end{gathered}$

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

$\begin{gathered} x_4=2\pi-\frac{\pi}{2} \\ x_4=\frac{3\pi}{2} \end{gathered}$

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:

$x=\frac{\pi}{3},\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{3}$