Question

Let f(x)= 1 + 3sec(x) and g(x)= -5. In the xy plane, what are the x coordinates of the points of intersection of the graphs f and g for 0 ≤ x ≤ 2π?

Answer 1

To find the x-coordinates of the points of intersection of the graphs of \( f(x) = 1 + 3\sec(x) \) and \( g(x) = -5 \), we need to set the two functions equal to each other:

\[
1 + 3\sec(x) = -5
\]

Subtracting 1 from both sides gives:

\[
3\sec(x) = -6
\]

Now, divide both sides by 3:

\[
\sec(x) = -2
\]

Recall that \(\sec(x) = \frac{1}{\cos(x)}\), so we can rewrite the equation as:

\[
\frac{1}{\cos(x)} = -2
\]

This implies:

\[
\cos(x) = -\frac{1}{2}
\]

Next, we find the values of \( x \) for which \( \cos(x) = -\frac{1}{2} \) in the interval \( 0 \leq x \leq 2\pi \). The cosine function is negative in the second and third quadrants.

The reference angle where \(\cos(x) = \frac{1}{2}\) is \( \frac{\pi}{3} \). Thus, the angles in the second and third quadrants where \(\cos(x) = -\frac{1}{2}\) are:

1. Second quadrant:
\[
x = \pi - \frac{\pi}{3} = \frac{2\pi}{3}
\]

2. Third quadrant:
\[
x = \pi + \frac{\pi}{3} = \frac{4\pi}{3}
\]

Therefore, the x-coordinates of the points of intersection of the graphs of \( f(x) \) and \( g(x) \) in the interval \( 0 \leq x \leq 2\pi \) are:

\[
\boxed{\frac{2\pi}{3}, \frac{4\pi}{3}}
\]