Problem 30 - Calculus

Question

Find a solution to the equation if possible. Give the answer in exact form and in decimal form. $$1=8 \tan (2 x+1)-3$$

Step-by-Step Solution

Answer

Exact: \( x = \frac{1}{2}(\tan^{-1}(1/2) - 1) + \frac{n\pi}{2} \). Approximation: \( x \approx -0.2682 + \frac{n\pi}{2}, \; n \in \mathbb{Z} \).

Step 1: Isolate the Trigonometric Function

Start by moving the constant term to the other side of the equation. We have the equation: \( 1 = 8 \tan(2x+1) - 3 \). Add 3 to both sides to get \( 4 = 8 \tan(2x+1) \).

Step 2: Simplify the Equation

Divide both sides of the equation by 8 to isolate the tangent function: \( \tan(2x+1) = \frac{4}{8} = \frac{1}{2} \).

Step 3: Solve for the Angle

Find the angle such that the tangent equals \( \frac{1}{2} \). This is done by using the inverse tangent function: \( 2x+1 = \tan^{-1}\left(\frac{1}{2}\right) \). An exact value solution isn't straightforward here, but we proceed using decimal approximation initially, then general solution.

Step 4: Solve for x

Subtract 1 from both sides: \( 2x = \tan^{-1}\left(\frac{1}{2}\right) - 1 \). Then divide by 2: \[ x = \frac{1}{2} (\tan^{-1}\left(\frac{1}{2}\right) - 1) \].

Step 5: General Solution

Remember that the tangent function is periodic with period \(\pi\). Thus, we include \( n\pi \) in the solution: \[ x = \frac{1}{2}(\tan^{-1}\left(\frac{1}{2}\right) - 1) + \frac{n\pi}{2}, \; n \in \mathbb{Z} \].

Step 6: Decimal Approximation

Using a calculator for \( \tan^{-1}\left(\frac{1}{2}\right) \), we find it is approximately 0.4636. Plug this in to get a numeric answer: \[ x \approx \frac{1}{2}(0.4636 - 1) = -0.2682 + \frac{n\pi}{2}, \; n \in \mathbb{Z} \].

Key Concepts

Tangent Function

When working with trigonometric equations, understanding the tangent function is essential. The tangent of an angle in a right triangle is the ratio of the opposite side to the adjacent side. It's defined as \(\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}\). Unlike sine and cosine, which only take values between -1 and 1, the tangent function can take any real number value.

Additionally, the tangent function has some unique properties:

• It's undefined whenever its input is an integer multiple of \(\frac{\pi}{2}\).
• It has vertical asymptotes at these points where the cosine of the angle is zero, as tangent is \(\frac{\sin(\theta)}{\cos(\theta)}\).

The tangent curve repeats every \(\pi\) radians, meaning it has a periodicity of \(\pi\). This is important when solving equations that involve the tangent function, as it allows us to account for all solutions within a given interval by adding multiples of \(\pi\).

Inverse Trigonometric Functions

Inverse trigonometric functions are very useful when you want to find the angle that corresponds to a given trigonometric value. In our context, the inverse tangent function, often denoted as \(\tan^{-1}(x)\), helps determine the angle whose tangent is \(x\). This process essentially "undoes" the tangent.

For example:

• If \(\tan(\theta) = \frac{1}{2}\), then \(\theta = \tan^{-1}\left(\frac{1}{2}\right)\).

This means you find an angle θ on the unit circle such that the tangent of \(\theta\) equals \(\frac{1}{2}\). Since tangent functions extend infinitely in the vertical direction due to their periodicity, there are an infinite number of solutions, but the inverse accessible through calculators or standard functions usually gives the value in the range \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).

Periodicity of Tangent

Periodicity refers to the repeating nature of a function. For the tangent function, this aspect is crucial because knowing it allows us to find all possible solutions to trigonometric equations. The tangent function is periodic with a period of \(\pi\). This means the function repeats its pattern every \(\pi\) units.

Because of this property, if you find one solution to an equation involving the tangent function, you can find others by adding integer multiples of \(\pi\) to that solution:

• If \(\tan(\theta) = c\), and \(\theta_0\) is a solution, then \(\theta = \theta_0 + n\pi\) also satisfies the equation for any integer \(n\).

In our given exercise, after finding the principal solution, you extend this with \( n\pi \) to account for all solutions across different cycles of the tangent function. This allows for complete representation of all possible angles satisfying the given trigonometric relation.