Problem 55 - A Graphical Approach to Precalculus with Limits

Question

Give solutions over the interval \([0,2 \pi)\) as approximations to the nearest hundredth when exact values cannot be determined. You may need to use the quadratic formula. Give approximate answers in Exercises \(59-64\) to the nearest tenth of a degree over the interval \(\left[0^{\circ}, 360^{\circ}\right)\) $$\tan ^{2} x+4 \tan x+2=0$$

Step-by-Step Solution

Answer

The approximate solutions are \( x \approx 108.4^{\circ}, 148.5^{\circ}, 288.4^{\circ}, 328.5^{\circ} \).

Step 1: Identify as a Quadratic Equation

Notice that the equation \( \tan^2 x + 4 \tan x + 2 = 0 \) can be treated as a quadratic equation in terms of \( \tan x \). This helps simplify the solving process.

Step 2: Apply the Quadratic Formula

Since the equation is quadratic in form, apply the quadratic formula: \( \tan x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 4 \), and \( c = 2 \). Substitute these values to find \( \tan x \).

Step 3: Calculate the Discriminant

Calculate the discriminant \( b^2 - 4ac = 4^2 - 4 \cdot 1 \cdot 2 = 16 - 8 = 8 \). The discriminant is positive, indicating two real solutions.

Step 4: Find the Roots using the Quadratic Formula

Substitute the discriminant and known values into the quadratic formula: \( \tan x = \frac{-4 \pm \sqrt{8}}{2} \). Simplify to find \( \tan x = -2 \pm \sqrt{2} \).

Step 5: Calculate Approximate Values for \( \tan x \)

Approximate \( \sqrt{2} \) as 1.41, which gives \( \tan x \approx -2 + 1.41 = -0.59 \) and \( \tan x \approx -2 - 1.41 = -3.41 \).

Step 6: Solve for \( x \) using Inverse Tangent

Use the inverse tangent to find \( x \). For \( \tan x = -0.59 \), \( x \approx 148.48^{\circ} \) and \( x \approx 328.48^{\circ} \). For \( \tan x = -3.41 \), \( x \approx 108.43^{\circ} \) and \( x \approx 288.43^{\circ} \).

Step 7: Wrap the Solutions into the Given Interval

Ensure all solutions \( x \) are within the specified interval \([0^{\circ}, 360^{\circ})\). The final angles are \( x \approx 108.4^{\circ}, 148.5^{\circ}, 288.4^{\circ}, 328.5^{\circ} \).

Key Concepts

Quadratic Formula

When faced with a trigonometric equation like \( \tan^2 x + 4 \tan x + 2 = 0 \), the quadratic formula becomes essential. This formula, \( \tan x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), helps find solutions to quadratic equations. In this context, substitute \( a = 1 \), \( b = 4 \), and \( c = 2 \). By calculating the discriminant \( b^2 - 4ac \), we check if the equation has real solutions.

• A positive discriminant, like 8 in this case, implies two real solutions.
• After determining the potential values of \( \tan x \), further steps will solve for the angle \( x \).

Using the quadratic formula simplifies solving complex trigonometric equations by treating them like simple quadratic ones. This technique is invaluable in providing approximate solutions where exact ones aren't possible.

Inverse Trigonometric Functions

Once \( \tan x \) is determined, convert these values into angles using inverse trigonometric functions, specifically the arctangent function. This function helps you find the angle whose tangent is a given number.

• For \( \tan x = -0.59 \), you'll find \( x \) using the inverse tangent function at approximately \( 148.48^{\circ} \) and \( 328.48^{\circ} \).
• Similarly, for \( \tan x = -3.41 \), the angles are \( 108.43^{\circ} \) and \( 288.43^{\circ} \).

Inverse trigonometric functions are fundamental in translating numeric values of tangent, sine, or cosine back into angle measures. This step transforms numerical answers into meaningful angles within specific intervals. Being skilled at using arctangent and other inverse functions is crucial for solving almost all trigonometric equations.

Approximation Techniques

In real-world applications and exams, you might need to approximate calculations. For expressions like \( \sqrt{2} \approx 1.41 \), approximation provides feasible solutions when precise calculations are complex or unnecessary.

• Approximating \( \tan x = -0.59 \) and \( \tan x = -3.41 \) involves simple arithmetic after obtaining values from the quadratic formula.
• Conversion of these tangential values into degrees further benefits from approximation, especially when technology limitations exist.

Approximation is especially useful in trigonometry for obtaining results that are quick to calculate while maintaining acceptable accuracy. Learning to balance between exact and approximated solutions helps you navigate exercises effectively, without getting bogged down in overly complex arithmetic.