Problem 26 - Precalculus with Calculus Previews

Question

Find all solutions of the given trigonometric equation if \(x\) is a real number and \(\theta\) is an angle measured in degrees. $$ 2 \sin ^{2} \theta+(2-\sqrt{3}) \sin \theta-\sqrt{3}=0 $$

Step-by-Step Solution

Answer

\( \theta = 21.45^\circ + 360^\circ n \) and \( \theta = 158.55^\circ + 360^\circ n \), where \( n \) is an integer.

Step 1: Identify the type of equation

The given equation \( 2 \sin^2 \theta + (2-\sqrt{3}) \sin \theta - \sqrt{3} = 0 \) is a quadratic equation in terms of \( \sin \theta \). We need to solve for \( \sin \theta \).

Step 2: Substitute variables into the quadratic formula

Let's denote \( \sin \theta = y \). Therefore, the equation becomes:\[ 2y^2 + (2-\sqrt{3})y - \sqrt{3} = 0. \] This implies \( a = 2, b = 2-\sqrt{3}, c = -\sqrt{3} \).

Step 3: Solve the quadratic equation using the quadratic formula

The quadratic formula is \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute the values of \( a, b, \) and \( c \): \[ y = \frac{-(2-\sqrt{3}) \pm \sqrt{(2-\sqrt{3})^2 - 4\cdot2\cdot(-\sqrt{3})}}{2\cdot2}. \]

Step 4: Simplify the quadratic formula

Calculate \( b^2 - 4ac \): \( (2 - \sqrt{3})^2 = 4 - 4\sqrt{3} + 3 = 7 - 4\sqrt{3} \) and \(-8(-\sqrt{3}) = 8\sqrt{3} \). Thus, \[ 7 - 4\sqrt{3} + 8\sqrt{3} = 7 + 4\sqrt{3}. \]Substitute back into the formula: \[ y = \frac{-(2-\sqrt{3}) \pm \sqrt{7 + 4\sqrt{3}}}{4}. \]

Step 5: Find numerical solutions for \(y\)

Calculate \( \sqrt{7 + 4\sqrt{3}} \) using a calculator to proceed with the division:Calculating, we find two approximate solutions, \( y_1 \approx 0.366 \) and \( y_2 \approx -2 \). Since \( y = \sin \theta \) and \( \sin \theta \) must be between -1 and 1, we discard \( y_2 = -2 \).

Step 6: Evaluate the angle for valid \(y\)

Since we found \( \sin \theta \approx 0.366 \), find \( \theta \) using the inverse sine function \( \theta = \sin^{-1}(0.366) \). Calculate it to get \( \theta \approx 21.45^\circ \). Additionally, since the sine function is positive in both the first and second quadrant, \( \theta \approx 180^\circ - 21.45^\circ = 158.55^\circ \).

Step 7: Write the general solution

The general solution for \( \sin \theta \approx 0.366 \) is given by:\[ \theta = 21.45^\circ + 360^\circ n \] and \[ \theta = 158.55^\circ + 360^\circ n \] for any integer \( n \). These angles represent the infinite set of solutions.

Key Concepts

Quadratic Equations in Trigonometry

In the exercise, we are working with a trigonometric equation that can be transformed into a quadratic form. The original equation is \( 2 \sin^2 \theta + (2-\sqrt{3}) \sin \theta - \sqrt{3} = 0 \). Recognizing that this is a quadratic equation in terms of \( \sin \theta \) is the first step.
A quadratic equation has the form \( ax^2 + bx + c = 0 \) and can be solved using the quadratic formula:

• \( a \) is the coefficient of \( x^2 \).
• \( b \) is the coefficient of \( x \).
• \( c \) is the constant term.
• The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

In our equation, we let \( \sin \theta = y \) to fit into the quadratic formula, resulting in \( a = 2 \), \( b = 2-\sqrt{3} \), and \( c = -\sqrt{3} \). This substitution simplifies solving trigonometric equations by treating them like traditional algebraic quadratic equations.

Inverse Trigonometric Functions

Once we find a solution for \( y = \sin \theta \), to determine \( \theta \), we must use inverse trigonometric functions. When \( y = \sin \theta \approx 0.366 \), to find \( \theta \), we use the inverse sine function:
\( \theta = \sin^{-1}(0.366) \).
Inverse trigonometric functions help us retrieve the angle given a specific sine, cosine, or tangent value.

• These functions include \( \sin^{-1} \), \( \cos^{-1} \), and \( \tan^{-1} \).
• The principal value of \( \sin^{-1} \) lies between \(-90^\circ\) and \(90^\circ\).

However, since the sine function is positive in both the first and second quadrants, we need to consider both corresponding angles. So, besides the principal angle \( \theta \approx 21.45^\circ \), we also find \( \theta \approx 180^\circ - 21.45^\circ = 158.55^\circ \). This comprehensive understanding allows for the complete set of solutions when using inverse trigonometric functions.

General Solution in Degrees

The general solution of a trigonometric equation accounts for all possible angles that satisfy the equation.
For the trigonometric equation \( 2 \sin^2 \theta + (2-\sqrt{3}) \sin \theta - \sqrt{3} = 0 \), after finding \( \theta \approx 21.45^\circ \) and \( 158.55^\circ \), we develop the general solution.
Trigonometric functions are periodic, meaning they repeat their values every specified interval (360 degrees for sine and cosine).

• The general solution in degrees is expressed as \( \theta = 21.45^\circ + 360^\circ n \) and \( \theta = 158.55^\circ + 360^\circ n \) where \( n \) is any integer.
• Each \( n \) increment represents one full rotation of the circle, bringing you back to an equivalent angle.

By including the variable \( n \), the general solution addresses the infinite nature of trigonometric solutions, conveying all possibilities within a periodic framework.