Problem 41 - Algebra and Trigonometry

Question

Involve trigonometric equations quadratic in form. Solve each equation on the interval \([0,2 \pi)\) $$ 2 \cos ^{2} x+3 \cos x+1=0 $$

Step-by-Step Solution

Answer

The solutions to the equation are \(x=2\pi/3, x=4\pi/3\) and \(x=\pi\)

Step 1: Change the equation to standard quadratic form.

We can notice that the given equation behaves similar to a quadratic equation. So let's replace \(\cos x\) with \(m\) for simplicity purposes. Hence, the equation turns into \(2m^2+3m+1=0\).

Step 2: Solve the quadratic equation.

Now we proceed to solve the quadratic equation using the quadratic formula \(m=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) where a=2, b=3, and c=1. Substituting these into the formula we get \(m=\frac{-3\pm\sqrt{3^2-4*2*1}}{2*2}=\frac{-3\pm\sqrt{1}}{4}\). Hence \(m_1=\frac{-3+1}{4}=-\frac{1}{2}\) and \(m_2=\frac{-3-1}{4}=-1\). Hence, the solutions to the quadratic equation are \(m_1=-\frac{1}{2}\) and \(m_2=-1\).

Step 3: Convert back to x.

Since we initially replaced \(\cos x\) with \(m\), we need to find the values of x that match these solutions. This involves two calculations: \(\cos x=-\frac{1}{2}\) and \(\cos x= -1\). The solutions to these within the interval \([0,2\pi)\) are \(x_1=2\pi/3, x_2=4\pi/3\) for the first equation and \(x_3=\pi\) for the second equation.

Key Concepts

Quadratic Equations

Quadratic equations are equations in the form of \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) represents an unknown variable. These equations are called 'quadratic' because the highest exponent of the variable \(x\) is 2, which is also referred to as the 'degree' of the equation.

Solving a quadratic equation typically involves finding the values of \(x\) that make the equation true. One effective method is using the quadratic formula:

• \(m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

This formula derives the roots of the equation by considering the coefficients \(a\), \(b\), and \(c\). The expression inside the square root, \(b^2 - 4ac\), is called the discriminant and determines the nature of the roots:

• If \(b^2 - 4ac > 0\), the equation has two real and distinct solutions.
• If \(b^2 - 4ac = 0\), the equation has exactly one real solution.
• If \(b^2 - 4ac < 0\), the equation has no real solutions.

In the context of trigonometric equations, replacing trigonometric expressions with a variable makes it easier to visualize and solve the equation.

Cosine Function

The cosine function is one of the fundamental trigonometric functions and is defined in the context of a right triangle or on the unit circle. In terms of a right triangle, \(\cos(\theta)\) is the ratio of the adjacent side to the hypotenuse. On the unit circle, \(\cos(\theta)\) represents the horizontal coordinate of a point on the circle corresponding to the angle \(\theta\).

The cosine function is periodic with a period of \(2\pi\), which means its values repeat after every \(2\pi\) interval. This property is crucial when solving trigonometric equations since it allows us to predict and identify multiple solutions based on the periodic nature of the function. The function ranges between \(-1\) and \(1\), hitting \(1\) at \(\theta = 0, 2\pi\) and \(-1\) at \(\theta = \pi\).When solving equations like \(\cos(x) = -\frac{1}{2}\) or \(\cos(x) = -1\), it's important to interpret the solutions within a given interval. For example:

• \(\cos(x) = -\frac{1}{2}\) occurs at \(x = 2\pi/3\) and \(x = 4\pi/3\) within the interval \([0, 2\pi)\).
• \(\cos(x) = -1\) is found at \(x = \pi\) in the same interval.

Interval Solutions

Interval solutions to trigonometric equations involve determining all potential solutions within a specified range or interval. In this case, the interval is \([0, 2\pi)\), which corresponds to one full rotation around the unit circle without including \(2\pi\) itself.

Finding solutions within an interval requires careful attention to the periodic properties of trigonometric functions like cosine. Because these functions repeat their values in a predictable pattern, solving the equation over an interval ensures that all valid solutions are found without duplicating them beyond the specified range.Here's a structured approach to finding these solutions:

• First, solve the equation algebraically and find potential roots or critical points of the function.
• Evaluate these solutions to check if they lie within the given interval.
• If the function is periodic (like cosine), consider the period to determine all solutions in the interval.

By following this approach, you can systematically find all possible solutions to trigonometric equations within any specified interval, ensuring completeness and accuracy in the solution set. This technique is invaluable in mathematical and real-world applications, where interpreting results within a limited scope is often necessary.