Problem 55 - Precalculus: Functions and Graphs

Question

Use inverse trigonometric functions to find the solutions of the equation that are in the given interval, and approximate the solutions to four decimal places. $$\begin{aligned} &2 \tan ^{2} t+9 \tan t+3=0\\\ &\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \end{aligned}$$

Step-by-Step Solution

Answer

The solutions are approximately \( t \approx -0.3403 \) and \( t \approx -1.3395 \) radians within the interval \((-\frac{\pi}{2}, \frac{\pi}{2})\).

Step 1: Identify the Trigonometric Equation

The problem gives us the equation \( 2 \tan^2 t + 9 \tan t + 3 = 0 \). We're asked to find solutions for \( t \) within the interval \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \).

Step 2: Substitute \( u = \tan t \)

To simplify the problem, let \( u = \tan t \). Then, the equation becomes \( 2u^2 + 9u + 3 = 0 \). Now, we need to find values of \( u \) that satisfy this quadratic equation.

Step 3: Solve the Quadratic Equation

Use the quadratic formula, \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = 9 \), and \( c = 3 \).Calculate the discriminant: \[ b^2 - 4ac = 9^2 - 4 \cdot 2 \cdot 3 = 81 - 24 = 57 \]Thus, \[ u = \frac{-9 \pm \sqrt{57}}{4} \]

Step 4: Compute the Roots

Calculate the roots:For positive root: \[ u_1 = \frac{-9 + \sqrt{57}}{4} \approx -0.3542 \]For negative root: \[ u_2 = \frac{-9 - \sqrt{57}}{4} \approx -4.2458 \]

Step 5: Find the Corresponding Angles

Now we substitute back to find \( t \) using \( t = \tan^{-1}(u) \):For \( u_1 = -0.3542 \), find\[ t_1 = \tan^{-1}(-0.3542) \approx -0.3403 \] radians.For \( u_2 = -4.2458 \), find \[ t_2 = \tan^{-1}(-4.2458) \approx -1.3395 \] radians.

Key Concepts

Quadratic Equations

In many mathematical exercises, quadratic equations often appear as foundational elements. A quadratic equation in its standard form is expressed as \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are coefficients with \( a eq 0 \). Solving quadratic equations can be done using various methods, such as factoring, completing the square, or using the quadratic formula.

• Quadratic equations can take different forms, including pure quadratics where \( b = 0 \).
• The solutions to these equations are referred to as roots and can be real or complex numbers.
• The nature of the roots is determined by the discriminant \( b^2 - 4ac \).

In our exercise, we transformed a trigonometric equation into a quadratic equation by setting \( u = \tan t \). This produced \( 2u^2 + 9u + 3 = 0 \), which is the standard form for a quadratic equation. By applying the quadratic formula, we found the roots \( u_1 \) and \( u_2 \), which are necessary to further solve for \( t \) in the trigonometric problem.

Trigonometric Identities

Trigonometric identities are fundamental in simplifying and solving trigonometric equations. These identities express relationships between the trigonometric functions: sine, cosine, tangent, cosecant, secant, and cotangent.

• The inverse trigonometric functions, such as \( \tan^{-1}(x) \), are used to find the angle whose trigonometric function equals a given number.
• By knowing the identities, you can convert complex trigonometric expressions into simpler forms.

In the exercise, the trigonometric identity used was the tangent function, \( \tan t \), which was substituted with \( u \) to form a quadratic equation. After finding the roots of the quadratic, using \( t = \tan^{-1}(u) \) allowed us to determine the angles \( t_1 \) and \( t_2 \) which satisfy the original equation in the specific interval.

Interval Solutions

Finding solutions within a specific interval is a crucial step in solving trigonometric equations, particularly because trigonometric functions are periodic. This means they repeat their values in certain intervals.

• The interval \((-\pi/2, \pi/2)\) aligns with the principal value range of the inverse tangent function, \( \tan^{-1} \), ensuring unique and meaningful solutions.
• It is essential to ensure that the solutions obtained fall within the specified interval for applicability and correctness.

In this problem, after determining the tangent values \( u_1 \) and \( u_2 \), we calculated the corresponding angles \( t_1 \) and \( t_2 \) using \( \tan^{-1} \). Checking these angles within the interval \((-\pi/2, \pi/2)\) confirmed that they were valid solutions. This method also aids in verifying the correctness of the solutions by observing their placement on the function's graph.