Problem 56 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 53-64: 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. \(3 \sin ^{2} t+7 \sin t+3=0 ; \quad\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)

Step-by-Step Solution

Answer

The solution is \( t \approx -0.6021 \).

Step 1: Recognize and Recast the Equation

The given equation is a quadratic in form: \( 3 \sin^2 t + 7 \sin t + 3 = 0 \). Start by letting \( x = \sin t \). This transforms the equation into a standard quadratic equation: \( 3x^2 + 7x + 3 = 0 \).

Step 2: Solve the Quadratic Equation

We solve \( 3x^2 + 7x + 3 = 0 \) using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)where \( a = 3, b = 7, c = 3 \).Calculate the discriminant: \( b^2 - 4ac = 49 - 36 = 13 \)The solutions for \( x \) are:\[x = \frac{-7 \pm \sqrt{13}}{6}\]

Step 3: Calculate the Values of x

Calculate the two solutions for \( x \):\( x_1 = \frac{-7 + \sqrt{13}}{6} \approx -0.5666 \)\( x_2 = \frac{-7 - \sqrt{13}}{6} \approx -1.7666 \)Since \( \sin t \) can only be in the range \([-1, 1]\), only \( x_1 = -0.5666 \) is valid for \( t \).

Step 4: Use the Inverse Sine Function

Find \( t \) by computing: \( t = \sin^{-1}(-0.5666) \).Approximate \( t \approx -0.6021 \).This value of \( t \) is in the required interval \([-\frac{\pi}{2}, \frac{\pi}{2}]\).

Step 5: Verify the Solution

Substitute \( t \approx -0.6021 \) back into the original trigonometric equation to verify:\( \sin t \approx -0.5666 \) is a solution for the original equation, confirming our calculations are accurate given the constraints.

Key Concepts

Quadratic Equations

Quadratic equations are algebraic expressions of degree two, which typically take the form \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are constants, and \( x \) represents the variable. To solve them, we often use the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This formula provides solutions for any quadratic equation, assuming the discriminant \( b^2 - 4ac \) is non-negative. The process involves substituting the values of \( a \), \( b \), and \( c \) into the formula and calculating the two possible values of \( x \). These are called roots or solutions of the quadratic equation.
Whenever dealing with quadratic forms that involve trigonometric functions, like "\( \sin t\)", substituting a variable simplifies solving using familiar algebraic methods.

Sine Function

The sine function is one of the fundamental trigonometric functions, depicted as \( \sin \theta \), where \( \theta \) is the angle, usually measured in radians. It relates to the y-coordinate on a unit circle, a representation of all angles as a circle with a radius of 1. Notably, the sine function:

• Varies between -1 and 1 throughout its range.
• Repeats every \( 2\pi \) radians, meaning it's periodic.

Sine plays a crucial role in modeling periodic phenomena, but its values are key when translating other mathematical expressions into solutions involving angles.
In solving equations like \( 3 \sin^2 t + 7 \sin t + 3 = 0 \), understanding this function’s properties is vital, especially to ensure solutions like \( \ -0.5666 \) remain within this range.

Trigonometric Intervals

Trigonometric intervals define ranges of angle measures, often expressed in radians. When solving trigonometric equations, it's crucial to find solutions within these specified intervals:

• Common trigonometric intervals include \([0, 2\pi]\), \([0, \pi]\), and \([-\frac{\pi}{2}, \frac{\pi}{2}]\).
• These intervals designate parts of the function where properties such as inverse calculations remain valid or simplified.

In the exercise, the interval \([-\frac{\pi}{2}, \frac{\pi}{2}]\) is given, which is used for calculating the inverse sine values. This interval allows solutions to be mapped back to the unit circle, offering a single, consistent value for the inverse function.

Inverse Sine

The inverse sine function, denoted as \( \sin^{-1}(x) \) or \( \arcsin(x) \), helps determine the angle \( \theta \) whose sine is \( x \). It's vital when you have the sine value and need to find the corresponding angle. Key points include:

• The domain of inverse sine is restricted to \([-1, 1]\), aligning with the range of the typical sine function.
• The range of the \( \arcsin \) function itself is \([-\frac{\pi}{2}, \frac{\pi}{2}]\), ensuring angles reflect the interval used in periodic calculations.

Applying this to \( x_1 = -0.5666 \), we find \( t \) such that \( t \approx -0.6021 \), a value fitting within the prescribed interval.
This method is instrumental when verifying trigonometric identities or confirming the accuracy of earlier calculations.