https://file.aiquickdraw.com/mathbot/file/1726457787090zdem79bu.jpg

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.

Solve the trigonometric equation for all values 0 < x < 10. sec(1/2 * x) + 1 = 0

Learn how to solve the trigonometric equation sec(1/2 * x) + 1 = 0 for values of x between 0 and 10. This problem involves secant and cosine functions and includes key trigonometric concepts. Suitable for Grades 10-12.

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