The trigonometric equation to solve is:

$\tan(2x) + 1 = 0$

We need to find all solutions for $x$ within the interval $0 < x \leq 2\pi$.

Step 1: Isolate the tangent function

Start by isolating the tangent function:

$\tan(2x) = -1$

Step 2: Find the general solution for $2x$

The tangent function equals $-1$ at the following angles:

$2x = \frac{3\pi}{4} + n\pi \quad \text{where } n \text{ is an integer}$

Step 3: Solve for $x$

To find $x$, divide both sides by 2:

$x = \frac{3\pi}{8} + \frac{n\pi}{2}$

Step 4: Determine $n$ such that $0 < x \leq 2\pi$

Substitute into the interval $0 < x \leq 2\pi$:

$0 < \frac{3\pi}{8} + \frac{n\pi}{2} \leq 2\pi$

Now, solve for $n$ by considering both ends of the inequality.

1. Lower bound: $\frac{3\pi}{8} + \frac{n\pi}{2} > 0$ Simplifying, this is always true for any $n$.

2. Upper bound: $\frac{3\pi}{8} + \frac{n\pi}{2} \leq 2\pi$ Subtract $\frac{3\pi}{8}$ from both sides: $\frac{n\pi}{2} \leq 2\pi - \frac{3\pi}{8} = \frac{13\pi}{8}$ Multiply by 2: $n\pi \leq \frac{13\pi}{4}$ Divide by $\pi$: $n \leq \frac{13}{4} = 3.25$ So, $n$ can be 0, 1, 2, or 3.

Step 5: Find the specific solutions

Substitute these values of $n$ into $x = \frac{3\pi}{8} + \frac{n\pi}{2}$:

• For $n = 0$: $x = \frac{3\pi}{8}$

• For $n = 1$: $x = \frac{3\pi}{8} + \frac{\pi}{2} = \frac{7\pi}{8}$

• For $n = 2$: $x = \frac{3\pi}{8} + \frac{\pi}{2} \times 2 = \frac{11\pi}{8}$

• For $n = 3$: $x = \frac{3\pi}{8} + \frac{\pi}{2} \times 3 = \frac{15\pi}{8}$

Final Answer:

The solutions to the equation $\tan(2x) + 1 = 0$ in the interval $0 < x \leq 2\pi$ are:

$x = \frac{3\pi}{8}, \quad \frac{7\pi}{8}, \quad \frac{11\pi}{8}, \quad \frac{15\pi}{8}$

Would you like more details, or do you have any other questions?

Related Questions:

1. What are the solutions to $\tan(kx) + c = 0$ for different constants $k$ and $c$?
2. How do you solve the equation $\sin(2x) = -1$ in the interval $0 < x \leq 2\pi$?
3. What is the period of the function $\tan(2x)$ and how does it affect the number of solutions?
4. How can you verify trigonometric solutions using a unit circle?
5. What changes if the equation involves $\cot(2x)$ instead of $\tan(2x)$?

Tip:

When working with trigonometric equations, always consider the periodicity of the function to determine all possible solutions within the given interval.