Problem 56 - Algebra and Trigonometry

Question

(a) Find all solutions of the equation. (b) Use a calculator to solve the equation in the interval \([0,2 \pi),\) correct to five decimal places. $$\tan ^{4} x-13 \tan ^{2} x+36=0$$

Step-by-Step Solution

Answer

The solutions in the interval \([0,2\pi)\) are approximately: 1.10715, 2.03444, 1.24905, 1.89255, 4.24874, 5.17603, 4.39064, and 5.03414.

Step 1: Rewrite the Equation

Notice that \( \tan^4 x - 13 \tan^2 x + 36 = 0 \) can be written in terms of a quadratic form if we make the substitution \( u = \tan^2 x \). This gives us the new equation: \( u^2 - 13u + 36 = 0 \).

Step 2: Solve the Quadratic Equation

Using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \(a = 1\), \(b = -13\), and \(c = 36\), we find the solutions for \( u \).\[ u = \frac{13 \pm \sqrt{(-13)^2 - 4 \cdot 1 \cdot 36}}{2 \cdot 1} \]\[ u = \frac{13 \pm \sqrt{169 - 144}}{2} \]\[ u = \frac{13 \pm \sqrt{25}}{2} \]\[ u = \frac{13 \pm 5}{2} \]So, the solutions for \( u \) are \( u = 9 \) and \( u = 4 \).

Step 3: Back Substitute for \(\tan^2 x\)

Substitute back to find \( \tan^2 x \): 1. If \( u = 9 \), then \( \tan^2 x = 9 \).2. If \( u = 4 \), then \( \tan^2 x = 4 \).So, we have two sets of equations: \( \tan x = \pm 3 \) and \( \tan x = \pm 2 \).

Step 4: Solve for \(x\)

Using the calculator, find the values of \( x \) such that:1. For \( \tan x = 3 \), \( x \approx 1.24905 \ \text{and} \ x \approx 1.24905 + \pi \).2. For \( \tan x = -3 \), \( x \approx 1.89255 \ \text{and} \ x \approx 1.89255 + \pi \).3. For \( \tan x = 2 \), \( x \approx 1.10715 \ \text{and} \ x \approx 1.10715 + \pi \).4. For \( \tan x = -2 \), \( x \approx 2.03444 \ \text{and} \ x \approx 2.03444 + \pi \).

Step 5: Confirm Solutions Within the Interval

Given the interval \([0, 2\pi)\), check and confirm the solutions:- For \( \tan x = 3 \), solutions are \( 1.24905 \) and \( 4.39064 \). - For \( \tan x = -3 \), solutions are \( 1.89255 \) and \( 5.03414 \).- For \( \tan x = 2 \), solutions are \( 1.10715 \) and \( 4.24874 \).- For \( \tan x = -2 \), solutions are \( 2.03444 \) and \( 5.17603 \).These results fit within the interval.

Key Concepts

Understanding the Tangent Function

The tangent function, denoted as \( \tan(x) \), is one of the primary trigonometric functions and is defined as the ratio of the sine and cosine functions: \( \tan(x) = \frac{\sin(x)}{\cos(x)} \). It is important to understand that the tangent function repeats every \( \pi \) units, which is known as its period. This means that the graph of \( \tan(x) \) looks the same from \( x \) to \( x + \pi \), giving it a unique and repeating pattern.
Unlike sine and cosine, the tangent function has vertical asymptotes where the cosine equals zero. These occur at \( x = \frac{\pi}{2} + k\pi \), with \( k \) being any integer. As a result, the function can "blow up" to infinity when approaching these points. Recognizing these characteristics helps us understand the behavior of the tangent function and how equations involving \( \tan(x) \) are solved.
This concept is key when dealing with trigonometric equations, as it allows us to identify the solutions at specific intervals, understanding how they repeat and where they become undefined.

Transforming into Quadratic Equations

Quadratic equations are polynomial equations of the form \( ax^2 + bx + c = 0 \). Often, complex trigonometric equations can be rewritten in a quadratic form to simplify the solving process. For example, if you have an equation like \( \tan^4 x - 13 \tan^2 x + 36 = 0 \), you can use substitution to transform it.
By letting \( u = \tan^2 x \), the equation becomes \( u^2 - 13u + 36 = 0 \), which is a more standard quadratic equation. Solving this using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) provides the values for \( u \), which can then be substituted back to solve for \( \tan(x) \).
This technique highlights the power of substitution in simplifying complex problems, providing a clear path to finding solutions by temporarily altering the equation's variables.

Finding Interval Solutions

Once the quadratic equation has been solved and the values of \( an^2 x \) determined, the real challenge is to find the interval solutions that match the given range, such as \([0, 2\pi)\). A solution must lie within this interval to be valid.
The tangent function's periodicity means that solutions can repeat every \( \pi \) units, so once a solution for \( x \) is found, additional solutions within \([0, 2\pi)\) can be identified by adding or subtracting \( \pi \).
This step confirms if each calculated angle falls within the appropriate range. If solutions fall outside, they can be adjusted by adding or subtracting \( \pi \) multiples until they are positioned correctly. Confirming the correct placement within the interval is crucial, as it ensures that the solutions are accurate and comply with the constraints.

Using Calculators Efficiently

Calculators are indispensable tools when solving trigonometric equations, especially when high precision is required, like finding values correct to five decimal places. The calculator can rapidly compute the values of \( x \) for different angles when \( \tan(x) = k \) given specific values such as 3, -3, 2, and -2 in the discussed problem.
To leverage a calculator effectively, you should:

• Ensure that the mode is set to either radians or degrees, depending on the context of the problem.
• Use functions like \( \tan^{-1} \) to find principal solutions and to determine the angle whose tangent matches the given value.
• Accurately perform successive additions of \( \pi \) to find solutions within specific intervals.
• Check the solutions by substituting back into the original equation to confirm they satisfy it.

With these steps, the calculator becomes a reliable ally, providing precise and efficient solutions to even the most challenging trigonometric equations.