Problem 18 - Algebra 2 and Trigonometry

Question

In \(15-20\) , find, to the nearest hundredth of a radian, the values of \(\theta\) in the interval \(0 \leq \theta<2 \pi\) that satisfy the equation. $$ 3 \sin ^{2} \theta+7 \sin \theta+2=0 $$

Step-by-Step Solution

Answer

\( \theta \approx 3.48 \) and \( 5.94 \) radians.

Step 1: Identify the Trigonometric Substitution

We have the equation \(3 \sin^2 \theta + 7 \sin \theta + 2 = 0\). Notice that this is a quadratic in form, where \( \sin \theta \) is the variable. We can use a substitution to make it easier to solve.

Step 2: Let \( x = \sin \theta \)

Substitute \( x \) for \( \sin \theta \), transforming the equation into a standard quadratic equation: \[ 3x^2 + 7x + 2 = 0 \]

Step 3: Solve the Quadratic Equation

The quadratic equation \( 3x^2 + 7x + 2 = 0 \) can be solved using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 3 \), \( b = 7 \), and \( c = 2 \).

Step 4: Calculate the Discriminant

Calculate the discriminant \( b^2 - 4ac \):\[ 7^2 - 4 \times 3 \times 2 = 49 - 24 = 25 \]

Step 5: Solve for \( x \)

Substitute the values into the quadratic formula:\[ x = \frac{-7 \pm \sqrt{25}}{6} \]This gives us two solutions:\[ x = \frac{-7 + 5}{6} = -\frac{1}{3} \] and \[ x = \frac{-7 - 5}{6} = -2 \]

Step 6: Consider the Validity of Solutions

Since \( \sin \theta \) must be between -1 and 1, \( x = -2 \) is not valid. Thus, \( x = \sin \theta = -\frac{1}{3} \) is the only valid solution.

Step 7: Solve for \( \theta \)

Find \( \theta \) such that \( \sin \theta = -\frac{1}{3} \). Use the inverse sine function:\[ \theta = \sin^{-1}(-\frac{1}{3}) \]However, be cautious about the interval, as sine is negative in the third and fourth quadrants.

Step 8: Calculate \( \theta \) Values

Calculate \( \theta = \sin^{-1}(-\frac{1}{3}) \) which approximately equals -0.3398 radians. Adjust this for the interval:- For the third quadrant, \( \theta = \pi + 0.3398 \approx 3.48 \) radians.- For the fourth quadrant, \( \theta = 2\pi - 0.3398 \approx 5.94 \) radians.

Key Concepts

Quadratic Equations

Quadratic equations are expressions in the form of \( ax^2 + bx + c = 0 \), where \( a eq 0 \). To solve these equations, you can use multiple methods: factoring, completing the square, or the quadratic formula. In our exercise, we have a trigonometric equation that can be transformed into a quadratic form. By thinking of \( \sin \theta \) as \( x \), the problem becomes a quadratic equation: \( 3x^2 + 7x + 2 = 0 \). Once in quadratic form, the quadratic formula comes in handy:

• The quadratic formula is: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
• The discriminant, \( b^2 - 4ac \), helps determine the nature of the roots.

In this problem, the discriminant turns out to be positive, indicating two real solutions. After solving, we discard any solutions that do not fall within the acceptable range for the sine function.

Sine Function

The sine function is a crucial element in trigonometry, relating angles to ratios of side lengths in a right triangle. Sine of an angle \( \theta \), denoted as \( \sin \theta \), is the opposite side over the hypotenuse in a right triangle.

• The range of \( \sin \theta \) is \([-1, 1]\).
• It is periodic with a period of \( 2\pi \), meaning it repeats every \( 2\pi \) radians.

Sine values are unique for a given angle \( \theta \) within this interval. For our exercise, the function is negatively valued in the third and fourth quadrants, which helps when finding appropriate angles for the solutions. Recognizing where the sine function is positive or negative based on the quadrant is key in solving equations like this one.

Inverse Trigonometric Functions

Inverse trigonometric functions reverse the process of regular trigonometric functions. For sine, the inverse is \( \sin^{-1}(x) \), also known as arcsine.

• \( \sin^{-1}(x) \) undoes the sine function, giving an angle from a ratio.
• The typical range for \( \sin^{-1} \) is \([-\frac{\pi}{2}, \frac{\pi}{2}]\).

In the context of the exercise, after finding \( \sin \theta = -\frac{1}{3} \), we apply \( \sin^{-1} \) to find \( \theta \). Since sine is negative for \(-\frac{1}{3}\), we adjust by adding \( \pi \) and \( 2\pi \) to find the angles in the appropriate quadrants. The solutions must still reside within the interval \([0, 2\pi)\).

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for all values of the involving angles. Understanding these allows simplification and manipulation of trigonometric expressions.

• Basic identities like \( \sin^2 \theta + \cos^2 \theta = 1 \) are foundational.
• Identities can also help solve more complex trigonometric equations by rewriting parts of the equation in useful forms.

Although not heavily emphasized in this particular problem, familiarizing yourself with other identities will enhance your ability to handle various trigonometric equations and solve them more efficiently. Recognizing patterns and using these identities strategically can significantly simplify the problem-solving process.