Problem 67 - Algebra and Trigonometry

Question

Use an identity to solve each equation on the interval \([0,2 \pi)\) $$ 4 \cos ^{2} x=5-4 \sin x $$

Step-by-Step Solution

Answer

The solutions to the equation are \( x = \frac{3\pi}{2} \), \( x = \frac{\pi}{6} \), or \( x = \frac{5\pi}{6} \) on the interval \([0,2 \pi)\)

Step 1: Rewrite in terms of sine

To better handle the problem, rewrite the given equation in terms of sine using the Pythagorean identity. Recall that the Pythagorean Identity \(\cos ^{2} x = 1-\sin ^{2} x\), substituting into the given equation it becomes: \(4*(1-\sin^{2} x)=5-4 \sin x\). Expand this to get: \(4 - 4\sin^{2}x = 5 - 4\sin x\)

Step 2: Form a quadratic equation

Rearrange this equation to form a quadratic equation. The quadratic equation is as follows: \(4\sin^{2}x + 4\sin x - 1 = 0\). It will be easier to solve the equation by dividing all terms by 4, this yields: \(\sin^{2}x + \sin x - 0.25 = 0\)

Step 3: Use Quadratic Formula

This equation can be solved using the quadratic formula \(\sin x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\). Substituting the given values \(\sin x = \frac{-1 \pm \sqrt{1 + 1}}{2}\). This gives two possible solutions: \( \sin x = -1 \) or \( \sin x = 0.5 \)

Step 4: Solve for x

Next, find the values of x in the domain [0,2\(\pi\)] that satisfy the equations \( \sin x = -1 \) or \( \sin x = 0.5 \). The solutions are \( x = \frac{3\pi}{2} \), \( x = \frac{\pi}{6} \), or \( x = \frac{5\pi}{6} \)

Key Concepts

Pythagorean Identity

The Pythagorean identity is a fundamental tool in trigonometry. It relates the sine and cosine of an angle and is given by \( \cos^2 x + \sin^2 x = 1 \). This identity allows us to express one trigonometric function in terms of the other.
In many trigonometric equations, using this identity can simplify the problem by reducing the number of trigonometrical terms.
For example, in our original equation, we used the identity \( \cos^2 x = 1 - \sin^2 x \) to express the equation entirely in terms of sine.
By substituting \( \cos^2 x \) with \( 1 - \sin^2 x \), you can transform complicated problems and pave the way for simpler algebraic manipulation. This tactic often helps to unify the terms and create a quadratic equation that can be solved using other techniques.

Quadratic Equation

A quadratic equation is a polynomial equation of degree two. It generally takes the form \( ax^2 + bx + c = 0 \). Within the realm of solving trigonometric equations, converting the equation into this quadratic form is crucial.
In our example, after using the Pythagorean identity, we were able to rearrange the terms into the following quadratic equation: - \( 4\sin^2 x + 4\sin x - 1 = 0 \).
This was simplified by dividing through by 4 to get: - \( \sin^2 x + \sin x - 0.25 = 0 \).
Identifying the quadratic equation form in trigonometric problems helps to leverage the Quadratic Formula for finding precise solutions. Quadratic equations can represent diverse scenarios beyond pure algebra, making them a powerful component of mathematical problem-solving.

Quadratic Formula

The Quadratic Formula is a versatile tool used for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). It is given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
This formula allows you to find the roots of any quadratic equation, provided the values of \( a \), \( b \), and \( c \).
In solving our trigonometric equation, we applied the Quadratic Formula using \( a = 1 \), \( b = 1 \), and \( c = -0.25 \).
After substitution, the formula yielded the solutions for \( \sin x = -1 \) and \( \sin x = 0.5 \).
Using the Quadratic Formula provides a systematic method to determine all possible solutions, making it a staple in problem-solving strategies, especially when factors are not evident.

Solving Trigonometric Equations

Solving trigonometric equations involves finding all angles that satisfy the given condition within a specified interval. Typically, the interval is \([0, 2\pi)\), representing one complete rotation around the unit circle.
In our exercise, once we found the potential values of \( \sin x \) using the Quadratic Formula, we needed to determine the values of \( x \) that correspond to these sine values.

• For \( \sin x = -1 \), the solution within our domain is \( x = \frac{3\pi}{2} \).
• For \( \sin x = 0.5 \), the possible solutions are \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \).

Understanding how to read the unit circle and identify corresponding angle values is integral to solving these problems. Each solution to \( \sin x = \text{value} \) corresponds to specific angles on the unit circle depending on which quadrants the angles lie within. Being familiar with this process allows you to solve a wide range of trigonometric equations confidently.