Problem 49 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 39-62: Find the solutions of the equation that are in the interval \([0,2 \pi\) ). $$ \sin ^{2} \theta+\sin \theta-6=0 $$

Step-by-Step Solution

Answer

There are no solutions in the interval \([0, 2\pi)\).

Step 1: Understand the Equation

The equation \( \sin^2 \theta + \sin \theta - 6 = 0 \) is a quadratic equation in form, with respect to \( \sin \theta \). \( \sin \theta \) acts as the variable here instead of \( \theta \). This means we can use factoring or the quadratic formula to solve it, treating \( \sin \theta \) like \( x \).

Step 2: Rewrite the Equation

Recognize that \( \sin^2 \theta \) can be written as \( (\sin \theta)^2 \). Let's rewrite the equation as \( (\sin \theta)^2 + \sin \theta - 6 = 0 \), making it more explicit that it's quadratic in terms of \( \sin \theta \).

Step 3: Apply the Quadratic Formula

The equation now resembles \( x^2 + x - 6 = 0 \). We apply the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 1 \), and \( c = -6 \).

Step 4: Calculate Discriminant

First calculate the discriminant: \( b^2 - 4ac = 1^2 - 4 \times 1 \times (-6) = 1 + 24 = 25 \). The discriminant is positive, indicating two real solutions.

Step 5: Solve for \( \sin \theta \)

Now plug the values into the quadratic formula: \( x = \frac{-1 \pm \sqrt{25}}{2} \). This yields \( x = \frac{-1 \pm 5}{2} \), giving us \( x_1 = 2 \) and \( x_2 = -3 \).

Step 6: Evaluate Solutions in the Context of Sine Function

The sine function only takes values between -1 and 1, so \( x_1 = 2 \) and \( x_2 = -3 \) are not valid solutions because they fall outside the range of the sine function. Therefore, there are no \( \theta \) values in the interval \([0, 2\pi)\) that satisfy the original equation.

Key Concepts

Quadratic Equation

A quadratic equation is a type of polynomial equation where the highest degree of the variable is squared. It typically takes the form:

• \( ax^2 + bx + c = 0 \)

In this expression, \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). The presence of the squared term makes these equations "quadratic." To solve a quadratic equation, you can use various methods like factoring, the quadratic formula, or completing the square. Solving it usually yields two solutions, known as the roots of the equation.
When solving a quadratic equation in terms of trigonometric functions, such as \( \sin^2 \theta + \sin \theta - 6 = 0 \), \( \sin \theta \) essentially plays the role of \( x \). Thus, you treat \( \sin \theta \) as a variable when applying methods to solve the quadratic equation.
Quadratic equations often appear in real-world problems, including those involving motion, area, and other physical phenomena.

Sine Function

The sine function is a fundamental building block in trigonometry. It relates the angle \( \theta \) of a right triangle to the ratio of the length of the side opposite that angle to the hypotenuse. The function is periodic, with a period of \( 2\pi \), meaning it repeats its values in intervals of \( 2\pi \).

• The function is defined for all real numbers.
• Its range is limited to the interval \([-1, 1]\).

When solving an equation containing a sine function, the possible values of the function are restricted by this interval. Thus, any prospective solution, such as those found in a quadratic equation, must fall within this range. For example, when solutions like \( x_1 = 2 \) and \( x_2 = -3 \) arise, they must be rejected as they fall outside of the allowable range of the sine function. Understanding this limitation is crucial when determining viable solutions.

Discriminant

The discriminant in a quadratic equation is a specific value obtained from the coefficients of the equation. It is found using the expression:

• \( b^2 - 4ac \)

The discriminant helps in determining the nature of the roots of the quadratic equation:

• If the discriminant is positive, the equation has two distinct real roots.
• If it is zero, there is exactly one real root, or in other terms, a repeated root.
• A negative discriminant suggests that the roots are complex or imaginary numbers.

In the given exercise, we calculated a discriminant of 25. This positive value indicates that two real roots exist. However, when considering the constraints of the sine function, we found that the solutions \( x_1 = 2 \) and \( x_2 = -3 \) are not possible within its domain. The discriminant is a powerful tool that gives insight into the possible solutions before they are calculated, providing a snapshot of what to expect.

Interval Solutions

Interval solutions refer to finding solutions within a specified range of values. In trigonometric equations, such as the one in this problem, the solutions are often sought within a specific interval like \([0, 2\pi)\).
This interval represents one full cycle of the sine function. When solving equations involving trigonometric functions, it's important to check that solutions fall within the prescribed interval.

• Solve the equation as you normally would.
• Verify that the results match the constraints of the trigonometric function.
• Ensure the solutions also lie within the given interval.

In the given exercise, despite the quadratic equation potentially having real roots, neither solution is valid because they don't match the allowable range of the sine function. Thus, there are no solutions within \([0, 2\pi)\). By understanding interval solutions, you ensure that the final answers are not only mathematically correct but also contextually appropriate within the problem's conditions.