Problem 46 - Algebra and Trigonometry

Question

Find all solutions of the equation in the interval \([0,2 \pi).\) $$2 \sin ^{2} x-\cos x=1$$

Step-by-Step Solution

Answer

The solutions are \( x = \frac{\pi}{3}, \frac{5\pi}{3}, \pi \).

Step 1: Use Trigonometric Identity

To solve the equation, start by using the identity \( \sin^2 x = 1 - \cos^2 x \). Substitute this identity into the equation \( 2 \sin^2 x - \cos x = 1 \) which gives: \[ 2(1 - \cos^2 x) - \cos x = 1 \]Simplifying, we get: \[ 2 - 2 \cos^2 x - \cos x = 1 \]

Step 2: Rearrange Equation

Continue simplifying by collecting all terms on one side of the equation to form a standard quadratic in terms of \( \cos x \): \[-2 \cos^2 x - \cos x + 2 = 1 \]Subtract 1 from both sides:\[ -2 \cos^2 x - \cos x + 1 = 0 \]Multiply the whole equation by \(-1\) for simplicity:\[ 2 \cos^2 x + \cos x - 1 = 0 \]

Step 3: Solve Quadratic Equation

The equation \( 2 \cos^2 x + \cos x - 1 = 0 \) is a quadratic equation in \( \cos x \). Let \( y = \cos x \), then replace \( \cos x \) with \( y \):\[ 2y^2 + y - 1 = 0 \]To solve this quadratic equation, use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = 1 \), and \( c = -1 \).

Step 4: Apply Quadratic Formula

Substitute the values \( a = 2 \), \( b = 1 \), and \( c = -1 \) into the quadratic formula:\[ y = \frac{-1 \pm \sqrt{1^2 - 4 \times 2 \times (-1)}}{2 \times 2} \]This simplifies to:\[ y = \frac{-1 \pm \sqrt{1 + 8}}{4} \]\[ y = \frac{-1 \pm \sqrt{9}}{4} \]\[ y = \frac{-1 \pm 3}{4} \]

Step 5: Find Values of \(y\)

Calculate the two possible values of \( y \): 1. \( y = \frac{-1 + 3}{4} = \frac{1}{2} \)2. \( y = \frac{-1 - 3}{4} = -1 \)These represent \( \cos x = \frac{1}{2} \) and \( \cos x = -1 \).

Step 6: Find \(x\) for \( \cos x = \frac{1}{2} \)

For \( \cos x = \frac{1}{2} \), recall from trigonometric values that this occurs at \[ x = \frac{\pi}{3}, \frac{5\pi}{3} \] within the interval \([0, 2\pi)\).

Step 7: Find \(x\) for \( \cos x = -1 \)

For \( \cos x = -1 \), this happens at\[ x = \pi \] in the interval \([0,2\pi)\).

Step 8: Solution Set

The solutions for the equation \( 2 \sin^2 x - \cos x = 1 \) in the interval \([0, 2\pi)\) are:\[ x = \frac{\pi}{3}, \frac{5\pi}{3}, \pi \]

Key Concepts

Trigonometric Identities

Trigonometric identities are essential tools in solving trigonometric equations. They allow us to transform and simplify complex expressions. In this exercise, we used the identity \( \sin^2 x = 1 - \cos^2 x \). This identity stems from the fundamental Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \). By replacing \( \sin^2 x \) with \( 1 - \cos^2 x \), we could manipulate the equation into a more manageable form, specifically a quadratic equation in terms of \( \cos x \).
These identities are powerful because they provide a way to interrelate the trigonometric functions \( \sin, \cos, \) and \( \tan \). Knowing how to use these identities can greatly simplify the process of solving trigonometric equations and analyzing functions.

Quadratic Equations

After applying the trigonometric identity, the equation was transformed into a quadratic equation: \( 2 \cos^2 x + \cos x - 1 = 0 \). Quadratic equations take the general form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants. Solving quadratic equations often involves using the quadratic formula:

• \( y = \cos x \)
• Substitute into: \( 2y^2 + y - 1 = 0 \)
• Quadratic formula: \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

By substituting \( a = 2 \), \( b = 1 \), and \( c = -1 \), we solve for \( y \). The quadratic formula is particularly useful when factoring is difficult or impossible. It provides a reliable method to find the roots for any quadratic equation.

Interval Notation

Interval notation is a concise way to express a range of numbers. It is frequently used in mathematics, especially in calculus and algebra, to specify the domain or solution set of an equation. In this problem, we express solutions within the interval \( [0, 2\pi) \), which includes numbers starting from 0 up to, but not including, \( 2\pi \).

• "[" or "]" at the ends include the boundary (closed interval).
• "(" or ")" at the ends exclude the boundary (open interval).

For this exercise, we used a semi-open interval: \[ [0, 2\pi) \]. This means the solution includes 0 but not \( 2\pi \). Using interval notation helps to clearly define the range within which the solution is relevant, avoiding any ambiguity.