The equation shown in the image is:

$\sec\left(\frac{1}{2}x\right) + 1 = 0$

We can solve this step by step.

Step 1: Isolate the secant function

Start by moving the $1$ to the other side of the equation:

$\sec\left(\frac{1}{2}x\right) = -1$

Step 2: Recall the definition of secant

The secant function is the reciprocal of the cosine function, so:

$\sec\left(\theta\right) = \frac{1}{\cos\left(\theta\right)}$

This implies:

$\frac{1}{\cos\left(\frac{1}{2}x\right)} = -1$

Step 3: Solve for cosine

To solve for the cosine, take the reciprocal of both sides:

$\cos\left(\frac{1}{2}x\right) = -1$

Step 4: Solve for $x$

Now, we need to solve for $x$ where $\cos\left(\frac{1}{2}x\right) = -1$. The cosine function equals $-1$ at angles of the form:

$\frac{1}{2}x = \pi + 2k\pi \quad \text{for} \quad k \in \mathbb{Z}$

Solving for $x$:

$x = 2\pi + 4k\pi$

So, $x = 2\pi(1 + 2k)$, where $k$ is an integer.

Step 5: Apply the constraint $0 < x < 10$

We now check which values of $x$ lie within the given interval $0 < x < 10$. Let’s calculate the specific values of $x$ based on different values of $k$:

• For $k = 0$: $x = 2\pi \approx 6.2832$
• For $k = 1$: $x = 6\pi \approx 18.8496$ (this is greater than 10, so it's out of range)

Thus, the only solution in the interval $0 < x < 10$ is:

$x = 2\pi \approx 6.2832$

The solution is $x = 2\pi$.

Would you like more details, or have any questions?

Here are five related questions to explore:

1. How would the solution change if the interval were different, such as $0 < x < 20$?
2. What is the period of the cosine function?
3. How does the secant function behave near its asymptotes?
4. Can you explain the behavior of the cosine function between $0$ and $2\pi$?
5. How do transformations of trigonometric functions (like scaling) affect their graphs?

Tip: Always consider the interval given in trigonometric equations to limit your solutions appropriately.