Problem 15 - A Graphical Approach to Precalculus with Limits

Question

Solve each equation for solutions over the interval \([0,2 \pi)\) by first solving for the trigonometric finction. Do not use a calculator. $$\sin ^{2} x-2 \sin x+1=0$$

Step-by-Step Solution

Answer

The solution to the equation is \( x = \frac{\pi}{2} \).

Step 1: Simplify the Equation

The given equation is \( \sin^2 x - 2 \sin x + 1 = 0 \). Notice that it resembles a quadratic equation in the form \( a^2 - 2a + 1 = (a-1)^2 \). Rewrite the equation as \((\sin x - 1)^2 = 0\).

Step 2: Solve for the Trigonometric Function

To solve \((\sin x - 1)^2 = 0\), set \( \sin x - 1 = 0 \). Solve for \( \sin x \) which gives \( \sin x = 1 \).

Step 3: Find the Solutions in the Given Interval

Next, we find all \( x \) in the interval \([0, 2\pi)\) where \( \sin x = 1 \). The sine function is equal to 1 at \( x = \frac{\pi}{2} \).

Step 4: Conclusion

Since \( \sin x = 1 \) only at \( x = \frac{\pi}{2} \) in the given interval, the solution set is \( x = \frac{\pi}{2} \).

Key Concepts

Understanding the Sine Function

The sine function, denoted as \( \sin x \), is one of the fundamental trigonometric functions. It relates the ratio of the length of the opposite side to the hypotenuse in a right triangle. More generally, for any angle \( x \) measured in radians, the sine function represents the y-coordinate of the corresponding point on the unit circle.
The sine function has several key properties:

• It is periodic with a period of \( 2\pi \), meaning that its values repeat every \( 2\pi \) units.
• It is defined for all real numbers, with values ranging from -1 to 1.
• It reaches its maximum value of 1 at \( x = \frac{\pi}{2} + 2k\pi \) for any integer \( k \).
• It reaches its minimum value of -1 at \( x = \frac{3\pi}{2} + 2k\pi \) for any integer \( k \).

Understanding these properties helps in solving trigonometric equations like the one given. The problem asks when \( \sin x = 1 \), which occurs when \( x = \frac{\pi}{2} \) within the interval \([0, 2\pi)\).

Transforming to Quadratic Equations

The original trigonometric equation \( \sin^2 x - 2 \sin x + 1 = 0 \) can be thought of in a similar way to quadratic equations. Quadratic equations typically take the form \( ax^2 + bx + c = 0 \). By substituting \( \sin x \) with a variable like \( u \), the equation looks like \( u^2 - 2u + 1 = 0 \).
Consider some important aspects of quadratic equations:

• They can sometimes be factored into simpler expressions, like in our problem \( (u - 1)^2 = 0 \).
• The solutions to the equation come from finding the values of \( u \) that make this factored form zero.
• In the context of trigonometric equations, substituting back the trigonometric function leads to solutions in terms of angles.

Factoring is a common method to solve quadratics, and in this case, it makes the equation much simpler, finding immediately that \( \sin x = 1 \). This ties back into understanding that trigonometric identities can often be expressed in quadratic forms.

Applying Interval Notation

Interval notation is a mathematical way of writing subsets of the real number line. Here, we're focusing on the interval \([0, 2\pi)\), which includes all real numbers \( x \) such that \( 0 \leq x < 2\pi \).
The characteristics of interval notation are:

• Closed intervals, such as \([a, b]\), include the endpoints \( a \) and \( b \); however, in this problem, the interval is half-open because it uses \([0, 2\pi)\), indicating that \( 0 \) is included while \( 2\pi \) is not.
• It provides a concise way to express a range of numbers without listing them.
• Interval notation is particularly useful for expressing solutions to problems where a specific range of solutions is required, like in trigonometric problems where angles are typically restricted to intervals from 0 to \( 2\pi \).

Using interval notation, we clearly indicate that we're seeking solutions that fit within the given range. This makes interpreting the mathematical problem's context and solutions straightforward.