Problem 67 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 63-68: Approximate, to the nearest 10', the solutions of the equation in the interval \(\left[0^{\circ}, 360^{\circ}\right)\). $$ 12 \sin ^{2} u-5 \sin u-2=0 $$

Step-by-Step Solution

Answer

Approximate solutions are \( 40^{\circ}, 140^{\circ}, 190^{\circ}, \) and \( 350^{\circ} \).

Step 1: Identify the equation form

The given equation is \( 12 \sin^2 u - 5 \sin u - 2 = 0 \). This equation is quadratic in terms of \( \sin u \), which allows us to treat \( \sin u \) like a variable (say \( x \)) and rewrite the equation as \( 12x^2 - 5x - 2 = 0 \).

Step 2: Solve the quadratic equation

Apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 12 \), \( b = -5 \), and \( c = -2 \). Substitute these values:\[x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(12)(-2)}}{2(12)}\]This simplifies to:\[x = \frac{5 \pm \sqrt{25 + 96}}{24} = \frac{5 \pm \sqrt{121}}{24} = \frac{5 \pm 11}{24}\]This results in two solutions: \( x = \frac{16}{24} = \frac{2}{3} \) and \( x = \frac{-6}{24} = -\frac{1}{4} \).

Step 3: Find angle solutions for each \(\sin u\) value

First, solve for \( \sin u = \frac{2}{3} \). The reference angle \( \theta \) satisfying \( \sin \theta = \frac{2}{3} \) is approximately \( 41.8^{\circ} \). In the interval \([0^{\circ}, 360^{\circ})\), the angles that satisfy \( \sin u = \frac{2}{3} \) are approximately \( 40^{\circ} \) and \( 140^{\circ} \).Next, solve for \( \sin u = -\frac{1}{4} \). The reference angle \( \theta \) satisfying \( \sin \theta = \frac{1}{4} \) is approximately \( 14.5^{\circ} \). In the interval \([0^{\circ}, 360^{\circ})\), the angles that satisfy \( \sin u = -\frac{1}{4} \) are approximately \( 190^{\circ} \) and \( 350^{\circ} \).

Step 4: Round solutions to the nearest degree

The rounded solutions within the given interval are approximately \( 40^{\circ} \), \( 140^{\circ} \), \( 190^{\circ} \), and \( 350^{\circ} \). Since the problem asks to round to the nearest 10 degrees, we round the angles to \( 40^{\circ}, 140^{\circ}, 190^{\circ}, \) and \( 350^{\circ} \).

Key Concepts

Quadratic Equations

Quadratic equations are fundamental in algebra and can appear in various mathematical contexts. A quadratic equation typically takes the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) represents the variable.
In the world of trigonometric equations, it's common to encounter quadratics where trigonometric functions appear as the "variable."
For instance, in the problem \( 12 \sin^2 u - 5 \sin u - 2 = 0 \), the equation is quadratic in terms of \( \sin u \).
This means you can use the familiar form to find solutions by substituting \( x \) with \( \sin u \).

• Apply the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with specific values for the equation's coefficients.
• Solving quadratics in this way helps in breaking down complex trigonometric problems into simpler algebraic ones.

Recognizing these forms is crucial for efficiently tackling and solving trigonometric equations.

Sine Function

The sine function is a core concept in trigonometry, representing the ratio of the length of the opposite side to the hypotenuse in a right-angled triangle. In the unit circle, the sine of an angle \( u \) corresponds to the y-coordinate of the point where the terminal side of the angle intersects the circle.
In our exercise, we encountered sine in the context of \( \sin u \), which plays a vital role in the trigonometric equation.
The sine function varies between -1 and 1, which is essential when solving trigonometric equations because only values within this range are possible solutions.

• For \( \sin u = \frac{2}{3} \): Since \( \frac{2}{3} \) lies within the acceptable interval, it can represent an angle.
• For \( \sin u = -\frac{1}{4} \): Again, this value is legitimate as it lies within the possible output range of the sine function.

Understanding the behavior and properties of the sine function is key to accurately solving trigonometric equations.

Angle Approximation

Angle approximation is a technique used to find a more approachable angle value when exact computation may not be straightforward. It is particularly useful when the solution can allow for approximations, such as rounding to the nearest degree or specific increment.
In trigonometry, and in the given exercise, exact angle solutions might not always come out cleanly.
The solutions may need to be approximated, especially when dealing with more specific intervals.
To find the approximate angles:

• Calculate the reference angle for each sine value, providing a baseline for other angle calculations.
• Using trigonometric principles, such as symmetry and angle identities within different quadrants, find all angles that correspond to the sine values provided by the original equation.
• Round to the nearest reasonable or required increment, in this case, the nearest 10° as specified by the problem.

Approximating angles in such a way simplifies difficult computations and offers practical solutions, especially within constrained intervals.

Interval Solutions

Interval solutions in trigonometry refer to finding all possible angle solutions within a specified range. This is crucial in many real-world applications where only angles within certain bounds are valid.
For this exercise, solutions need to be confined to the interval \([0^{\circ}, 360^{\circ})\). This means only considering angles within a complete circle but not inclusive of 360°.
When seeking interval solutions:

• Identify all possible angles that make the trigonometric equation true within the interval.
• Consider the periodic nature of trigonometric functions, often providing multiple valid solutions in a full 360° rotation.
• Acknowledge the quadrants and how particular functions behave in each, affecting the sign and thus the angle solutions.

Ensuring solutions fit within the given interval aids in providing correct and applicable results, aligning with the problem's requirements.