Problem 63 - Algebra and Trigonometry

Question

Use an identity to solve each equation on the interval \([0,2 \pi)\) $$ 2 \cos ^{2} x+\sin x-1=0 $$

Step-by-Step Solution

Answer

The equation \(2 \cos^2 x + \sin x - 1=0\) has no solutions within the interval \([0,2 \pi)\).

Step 1: Replace \(cos^2x\) using Pythagorean Identity

Use the Pythagorean identity to replace \(\cos^2x\). The equation becomes \(2(1 - \sin^2x) + \sin x - 1 = 0\).

Step 2: Simplify the equation

Simplify the expression to create a quadratic equation in terms of \(\sin x\). This yields: \(-2 \sin^2 x + \sin x - 1 = 0\).

Step 3: Re-order the equation

Re-order the equation to have positive leading coefficient. That is, \(2 \sin^2 x - \sin x + 1 = 0\).

Step 4: Solve for \(\sin x\)

Use the quadratic formula \((\sin x = [-b \pm \sqrt{b^2 - 4ac}] / 2a)\) to solve for \(\sin x\). This does not yield any real solutions as the discriminant of the quadratic is less than 0.

Step 5: Conclude the solution

As no real solutions emerged from the previous step, it can be concluded that the original equation, \(2 \cos^2 x + \sin x - 1=0\), has no solutions within the interval \([0,2 \pi)\).

Key Concepts

Pythagorean Identity

In trigonometry, the Pythagorean identity is a foundational concept that links the sine and cosine functions. It is expressed as \[ \cos^2x + \sin^2x = 1. \]This identity is incredibly useful because it allows us to express one trigonometric function in terms of another.

• If you know the sine of an angle, you can find its cosine, and vice versa.
• This relationship is derived from the Pythagorean theorem applied to a unit circle.

In solving trigonometric equations, this identity frequently simplifies problems by reducing the number of different trigonometric functions that appear in the equation. For example, in the original problem, we used this identity to express \( \cos^2x \) as \( 1 - \sin^2x \), facilitating the transformation of the equation into a quadratic form.

Solving Trigonometric Equations

Trigonometric equations are equations involving trigonometric functions like sine, cosine, or tangent. To solve them, the goal is to find all the angles (usually within a specific interval) that make the equation true.

• Initially, simplify the equation using known identities or algebraic manipulation.
• Convert the equation to a recognizable form, such as a quadratic equation, which can be easier to handle.
• Always consider the interval given in the problem to determine possible solutions.

In the original exercise, we manipulated the trigonometric function to a quadratic form in terms of \( \sin x \). The transition from a trigonometric equation to a simpler algebraic equation is a powerful technique in solving these problems.

Quadratic Equations

Quadratic equations take on the standard form \[ ax^2 + bx + c = 0, \]where \( a, b, \) and \( c \) are constants. The solutions to these equations can be found using the quadratic formula:\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}. \]This formula provides the roots of the equation, and whether these roots are real or complex depends on what is called the discriminant, \( b^2 - 4ac \).

• If the discriminant is positive, two distinct real solutions exist.
• If it is zero, a single real solution exists.
• If it is negative, solutions are complex, indicating no real solution.

In the provided exercise, we apply this formula to the quadratic in terms of \( \sin x \). However, the discriminant is negative, indicating that there are no real solutions for \( x \) within the specified interval.

Interval [0, 2π)

An interval is a range of values typically used to specify where solutions to an equation are valid. In trigonometric contexts, common intervals include \[ [0, 2\pi) \ ext{and} \ [0, \pi]. \]The interval \([0, 2\pi)\) represents one full rotation around the unit circle, accounting for all possible angles from 0 to just under \(2\pi\). This is an essential consideration when solving trigonometric equations because it limits which solutions are valid based on their angular positions.

• The endpoints of this interval indicate the beginning at \(0\) and just before \(2\pi\), the wrap-around point for a circle.
• Solutions should be evaluated to ensure they fall within this range.

In the context of our problem, verifying solutions against this interval ensures consistency and correctness in finding valid \(x\) values for the given trigonometric equation.