Problem 20 - 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. $$ 2 \sin ^{2} x-3 \sin x+1=0 $$

Step-by-Step Solution

Answer

The solutions are \(x = 30^\circ + 360^\circ n\), \(x = 150^\circ + 360^\circ n\), and \(x = 90^\circ + 360^\circ n\) for integer \(n\).

Step 1: Identify the Equation Type

The given equation \(2 \sin^2 x - 3 \sin x + 1 = 0\) is a quadratic equation in terms of \(\sin x\). Treat \(\sin x\) as a variable \(y\), so the equation becomes \(2y^2 - 3y + 1 = 0\).

Step 2: Factor the Quadratic Equation

Rewrite \(2y^2 - 3y + 1 = 0\) in factored form. We look for two numbers that multiply to \(2 \times 1 = 2\) and add to \(-3\). This leads us to \(-1\) and \(-2\). Factor the equation: \((2y-1)(y-1)=0\).

Step 3: Solve for the Variable \(y\)

Set each factor equal to zero to solve for \(y\):- For \((2y-1)=0\), solve to get \(y = \frac{1}{2}\).- For \((y-1)=0\), solve to get \(y = 1\).

Step 4: Solve for \(x\) in Terms of \(\sin x\)

Since \(y = \sin x\), we solve:- \(\sin x = \frac{1}{2}\): The solutions are \(x = 30^\circ + 360^\circ n\) and \(x = 150^\circ + 360^\circ n\), where \(n\) is an integer.- \(\sin x = 1\): The solutions are \(x = 90^\circ + 360^\circ n\), where \(n\) is an integer.

Key Concepts

Quadratic Equations

Quadratic equations are crucial in mathematics, as they pop up in various contexts, including trigonometric scenarios like the one in this problem. A quadratic equation is typically expressed in the form \(ax^2 + bx + c = 0\). The equation \(2 \sin^2 x - 3 \sin x + 1 = 0\) is a quadratic in disguise, with \( \sin x\) taking the place of the variable. This means you can treat \( \sin x\) as \(y\) and rewrite the equation as \(2y^2 - 3y + 1 = 0\). This kind of transformation is what helps to simplify and eventually solve the equation.

After recognizing it's quadratic, the solution involves factoring the equation. Here, the product \(2 \times 1 = 2\) and the sum \(-3\) help identify the factors. This leads us to the factorization \((2y - 1)(y - 1) = 0\). Solving these factors gives the potential values for \(y\), which in the trigonometric sense, are the possible values of \( \sin x\). Quadratic equations are often the key to solving more complex problems, especially when they are intertwined with other mathematical functions.

Sin Function

The sine function, denoted as \( \sin x\), is one of the fundamental functions in trigonometry. It describes the ratio of the opposite side to the hypotenuse in a right triangle. In this exercise, \( \sin x\) acts as a variable in a quadratic equation. To solve for \(x\), you must solve for \( \sin x\) first.

• When \( \sin x = \frac{1}{2}\), this corresponds to specific angles in the unit circle where sine equals this value.
• Similarly, \( \sin x = 1\) corresponds to another set of angles.

The nature of the sine function is periodic, with a period of 360 degrees, meaning its graph repeats every 360 degrees. That's why, when identifying solutions for \(x\), the general form includes adding \(360n\) (where \(n\) is an integer) to preserve this periodicity. This ensures all possible angle solutions are accounted for, as the sine wave covers all values in its range repeatedly.

Angle Measurement

Angle measurement in degrees is a fundamental aspect of trigonometry often used to express solutions involving trigonometric functions. This exercise provides solutions in degrees because angles can often be more intuitive in this measure, especially when working within the unit circle.

• Degrees are part of the circle representation, where a full circle is 360 degrees. The solutions of trigonometric equations often refer to positions on this circle.
• For example, \(x = 30^\circ\) or \(x = 150^\circ\) refer to specific points where the sine function attains the value of \( \frac{1}{2}\).
• Additionally, the solution \(x = 90^\circ\) is where \( \sin x\) reaches its peak value of 1.

Using degrees makes understanding trigonometric solutions easier. It allows us to quickly identify key angles on the unit circle, such as 30, 150, and 90 degrees, providing a visual and practical way to grasp and apply these mathematical concepts.