Problem 84 - Precalculus with Limits

Question

In Exercises 75 - 84, find all solutions of the equation in the interval \( \left[0,2\pi\right) \). \( \cos\left(x - \dfrac{\pi}{2}\right) + \sin^2 x = 0 \)

Step-by-Step Solution

Answer

The solutions to the equation \( \cos\left(x - \dfrac{\pi}{2}\right) + \sin^2 x = 0 \) in the interval [0,2π) are \(x = 0\), \(x = \pi\) and \(x = \frac{3\pi}{2}\)

Step 1: Substitute for Cosine Using Trigonometric Identity

We know that \( cos (A - B) = cos A cos B + sin A sin B \). Hence, substitute \( \cos\left(x - \frac{\pi}{2}\right) \) to get a new equation. The substitution using this identity will be \( \cos x \cos \left(\dfrac{\pi}{2}\right) + \sin x \sin\left(\dfrac{\pi}{2}\right) + \sin^2 x = 0 \). But \( \cos \left(\frac{\pi}{2}\right) = 0 \) and \( \sin \left(\frac{\pi}{2}\right) = 1 \). Substituting these values will give \( 0 + \sin x + \sin^2 x = 0 \)

Step 2: Rearrange the Equation

Re-arranging the terms to form a quadratic equation, we get \(\sin^2 x + \sin x = 0\). Which is a quadratic equation in sine of x.

Step 3: Apply Zero-product Property

Factorize the equation. This results in \(\sin x (\sin x + 1) = 0\). Now apply the zero-product property, which states that if \(ab = 0\), then either \(a = 0\) or \(b = 0\). This gives us two equations: \(\sin x = 0\) and \(\sin x = -1\)

Step 4: Solve for x

The solutions to \(\sin x = 0\) in the interval [0,2π) are \(x = 0\) and \(x = \pi\). The solution to \(\sin x = -1\) in the interval [0,2π) is \(x = \frac{3\pi}{2}\). Hence, the solutions to the original equation in the interval [0,2π) are \(x = 0\), \(x = \pi\) and \(x = \frac{3\pi}{2}\)

Key Concepts

Cosine Subtraction Formula

The cosine subtraction formula is an important trigonometric identity that helps simplify trigonometric expressions involving cosine of differences. It states that for any two angles, \( A \) and \( B \), the formula is given by:

• \( \cos(A - B) = \cos A \cos B + \sin A \sin B \)

This formula allows us to Express \( \cos(A - B) \) in terms of simple sines and cosines, making it vastly easier to work through complex trigonometric problems.
In our exercise, we used this formula to rewrite \( \cos\left(x - \frac{\pi}{2}\right) \) as:

• \( \cos x \cos \left(\frac{\pi}{2}\right) + \sin x \sin\left(\frac{\pi}{2}\right) \)

Given that \( \cos\left(\frac{\pi}{2}\right) = 0 \) and \( \sin\left(\frac{\pi}{2}\right) = 1 \), the expression simplifies to \( \sin x \). By using this transformation, we turned the original trigonometric equation into a simpler form, which was then easier to solve.

Trigonometric Identities

Trigonometric identities are equations that hold true for all values of the variable within a given domain. These identities often help simplify complex trigonometric expressions. Key ones include:

• Pythagorean identities like \( \sin^2 x + \cos^2 x = 1 \)
• Angle sum and difference identities, such as our earlier cosine subtraction formula
• Double angle and half-angle formulas

In this problem, trigonometric identities played a crucial role in simplifying the complex given equation, \( \cos(x - \frac{\pi}{2}) + \sin^2 x = 0 \). By substituting with identities, we effectively reduced the complexity.
Learning these identities can make solving trigonometric equations more straightforward. They remove obstacles by simplifying interactions between different trigonometric functions, paving the way for easier factorization and solution.

Zero-Product Property

The zero-product property is a basic principle of algebra that assists in solving equations with multiple factors. It states:

• If a product of factors is zero, then at least one of the factors must be zero. Mathematically, if \( ab = 0 \), then \( a = 0 \) or \( b = 0 \).

This property is frequently used when solving quadratic equations and was crucial in the given problem after trigonometric identities helped simplify the equation. Once rearranged, our equation took the form of:

• \( \sin^2 x + \sin x = 0 \), factorized into \( \sin x (\sin x + 1) = 0 \)

Applying the zero-product property, we derived the separate equations \( \sin x = 0 \) and \( \sin x + 1 = 0 \). Solving these gives us specific solutions within the defined interval. Remember, if you can factor an expression, the zero-product property is a go-to method for solving the resulting equations.