Problem 106 - Algebra and Trigonometry

Question

Use the most appropriate method to solve each equation on the interval \([0,2 \pi) .\) Use exact values where possible or give approximate solutions correct to four decimal places. $$ 2 \sin ^{2} x=2-3 \sin x $$

Step-by-Step Solution

Answer

The solutions to the equation \(2\sin^2x=2-3\sin x\) in the interval \([0,2\pi)\) are \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\).

Step 1: Analyzing and Rewriting the Equation

We will first analyze this trigonometric equation and rewrite it in the usual form which is often easier to handle. Given equation \( 2\sin^2x=2-3\sin x\) could be rewritten as \(2\sin^2x + 3\sin x - 2 = 0\).

Step 2: Factoring the Equation

We write this equation in a factored form. The equation becomes \((2\sin x - 1)(\sin x + 2) = 0\). This is derived from the multiplication rule of algebraic expressions.

Step 3: Finding solutions

Now we find solutions for the equation by setting each factor equal to zero. We obtain the two equations \(2\sin x - 1 = 0\) and \(\sin x + 2 = 0\). Solving these equations simplifies the solution into \(\sin x = \frac{1}{2}\) and \(\sin x = -2\).

Step 4: Using the Trigonometry Function Interval Property

Consider the interval property of trigonometry where \(-1 \leq \sin x \leq 1\). We see that -2 lies outside this range, so \(\sin x = -2\) can be discarded. Therefore only the solution \(\sin x = \frac{1}{2}\) holds.

Step 5: Calculating the Values of x

We recall that in the interval \([0,2\pi)\) the sine function equals \(\frac{1}{2}\) at the angles \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\). So we have two solutions for x in the given interval, these are \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\).

Key Concepts

Factoring Equations

Factoring is a powerful algebraic technique used to simplify equations and is especially useful in solving polynomial equations.
To factor the equation \(2\sin^2x + 3\sin x - 2 = 0\), we recognize it as a quadratic expression in terms of \(\sin x\).
For quadratics, factoring involves writing the expression as a product of simpler binomials.
This step is crucial as it allows us to apply the zero-product property, which states that if \(ab = 0\), then either \(a = 0\) or \(b = 0\).
In this case, \((2\sin x - 1)(\sin x + 2) = 0\), breaking it down into two different cases to solve for \(x\).
Factoring transforms an equation into a more manageable form and opens the path to finding solutions.

Trigonometric Identities

Trigonometric identities are equations that are true for all values of the involved variables. They help simplify trigonometric expressions and solve equations.
In this exercise, while direct trigonometric identities like the Pythagorean identity \(\sin^2x + \cos^2x = 1\) were not explicitly used, they remain beneficial tools.
IDentifying that the original equation \(2\sin^2x=2-3\sin x\) can be rearranged into a quadratic form, implicitly uses a fundamental understanding of trigonometry:

• Recognize the structure of equations.
• Manipulate using algebraic identities and transformations.
• Solve by factoring.

Grasping identities and their applications ensures that complex equations can be translated into simpler mathematical statements.

Interval Properties

The concept of interval properties in trigonometry refers to the idea that certain trigonometric functions have defined ranges of values.
For example, the sine and cosine functions range between -1 and 1.
This is an essential property that helps identify feasible solutions to trigonometric equations.
In our original problem, we used the interval \([-1, 1]\) for \(\sin x\) to discard the solution \(\sin x = -2\) because it lies outside the sine function's range.
This principle prevents false solutions from being considered and is critical for narrowing down potential answers in trigonometric problems.
Understanding interval properties ensures that solutions fall within the domain of the function, providing accurate results.

Exact Values in Trigonometry

Exact values are specific trigonometric ratios which remain constant for certain standard angles. These can be derived from geometric considerations like triangles or unit circles.
In the exercise, we found the sine function equals \(\frac{1}{2}\) at special angles \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\).

• They are in the interval \([0, 2\pi)\).
• Known as reference angles.
• Offer precise solutions without needing approximation.

Knowing these values by heart, or being able to derive them quickly, makes solving trigonometric equations much more efficient and accurate.