Find all solutions of the equation sec(2x) + 1 = 0 in the interval [0, 2π]. Write your answer in radians in terms of π.

Steps to Solve

1. Rearranging the Equation

Start by isolating the secant function in the equation:

[ \sec(2x) + 1 = 0 \quad \Rightarrow \quad \sec(2x) = -1 ]

2. Understanding Secant Function

Recall that the secant function is the reciprocal of the cosine function:

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

This means that:

[ \sec(2x) = -1 \quad \Rightarrow \quad \cos(2x) = -1 ]

3. Finding Angles Where Cosine is -1

The cosine function equals -1 at integer multiples of (\pi) where:

[ 2x = \pi + 2k\pi \quad \text{for } k \in \mathbb{Z} ]

This implies:

[ 2x = (2k + 1)\pi ]

4. Solving for (x)

Now, divide by 2 to solve for (x):

[ x = \frac{(2k + 1)\pi}{2} ]

5. Identifying Valid (k) Values within the Interval

Next, find values of (k) such that (x) lies within the interval ([0, 2\pi]):

[ 0 \leq \frac{(2k + 1)\pi}{2} \leq 2\pi ]

This simplifies to:

[ 0 \leq 2k + 1 \leq 4 \quad \Rightarrow \quad -1 \leq 2k \leq 3 ]

Thus, (k) can be (0) or (1):

• For (k = 0): [ x = \frac{(2 \cdot 0 + 1)\pi}{2} = \frac{\pi}{2} ]

• For (k = 1): [ x = \frac{(2 \cdot 1 + 1)\pi}{2} = \frac{3\pi}{2} ]

6. Listing All Solutions

The complete solutions for the equation in the interval ([0, 2\pi]) are:

[ x = \frac{\pi}{2}, \frac{3\pi}{2} ]