Question

If \(\cos x=\frac{2}{3}\) and \(\cos x=-\frac{1}{3}\) are the solutions for a trigonometric equation, what are the values of \(B\) and \(C\) if the equation is of the form \(9 \cos ^{2} x+B \cos x+C=0 ?\)

Short answer

# B = -3, C = -2

Step-by-step solution

- Set up the quadratic equation

Given the equation is of the form \( 9 \cos^{2} x + B \cos x + C = 0 \), substitute the solutions \( \cos x = \frac{2}{3} \) and \( \cos x = -\frac{1}{3} \).

- Write the equation with the known cosines

Since \( \cos x = \frac{2}{3} \) and \( \cos x = -\frac{1}{3} \) are solutions, the quadratic equation can be written as: \[ 9 \cos^{2} x + B \cos x + C = 9 (\cos x - \frac{2}{3})(\cos x + \frac{1}{3}) = 0 \]

- Expand the quadratic equation

Expand the equation \[ 9 (\cos x - \frac{2}{3})(\cos x + \frac{1}{3}) \] to identify the values of \( B \) and \( C \). Use distributive property: \[ 9(\cos^{2}x + \frac{1}{3}\cos x - \frac{2}{3}\cos x - \frac{2}{9}) \] Simplify inside the brackets: \[ 9(\cos^{2}x - \frac{1}{3}\cos x - \frac{2}{9}) \] Distribute the 9: \[ 9 \cos^{2}x - 3 \cos x - 2 = 0 \]

- Identify the coefficients B and C

From the expanded form, match the coefficients with \( 9 \cos^{2}x + B \cos x + C = 0 \) and determine that: \( B = -3 \) \( C = -2 \)

Key concepts

Trigonometric Identities

When working with trigonometric equations, it's vital to understand and use trigonometric identities. These identities are formulas that relate the angles and sides of triangles, and they help simplify trigonometric expressions.

Some of the key trigonometric identities include:

• Pythagorean Identity: \(\text{sin}^2 x + \text{cos}^2 x = 1\)
• Angle Sum and Difference Identities for sine and cosine
• Double Angle and Half Angle Identities

In our exercise, we deal specifically with the cosine function. Using these identities can convert complex expressions into simpler ones, making it easier to solve equations. Though not directly applied in this exercise, knowing these identities can be incredibly helpful in other trigonometric problems.

Cosine Function

The cosine function is one of the primary trigonometric functions. It is defined as the ratio of the adjacent side to the hypotenuse in a right-angled triangle. The cosine function is significant in many mathematical and applied contexts.

For our quadratic trigonometric equation, the variable inside the quadratic expression is actually \(\text{cos } x\). In this problem, you are given two solutions for \(\text{cos } x\), specifically \(\frac{2}{3}\) and -\(\frac{1}{3}\).

From these solutions, we can write the quadratic equation. Knowing the behavior of the cosine function, which varies between -1 and 1, aids in understanding why \(\frac{2}{3}\) and -\(\frac{1}{3}\) are valid solutions in the given context. Remembering that the cosine function is periodic and continuous helps in understanding the shape of its graph and where the solutions lie.

Quadratic Equations

Quadratic equations take the form: \ ax^2 + bx + c = 0 \. They are called quadratic because the highest exponent of the variable is two. Solving quadratic equations is a fundamental algebraic skill.

In our exercise, the quadratic equation is given in terms of \(\text{cos } x\), so it looks like \ 9 \text{cos}^2 x + B \text{cos} x + C = 0 \. The solutions to the quadratic equation are the values that \(\text{cos } x\) can take, which are \ \frac{2}{3} \ and \ -\frac{1}{3} \.

To find the values of B and C, we express the given solutions in terms of the original quadratic equation. Once expressed, we expand and simplify the product to identify the coefficients B and C. Understanding how to manipulate the equation and expand it is key to solving the problem correctly.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying algebraic expressions. This skill is crucial in solving equations, including trigonometric and quadratic equations.

In this exercise:

• We start with the given solutions for \(\text{cos } x\)
• We write the quadratic equation in its expanded form
• We identify coefficients by matching the expanded equation with the standard quadratic form

The distributive property is especially useful here. When expanding \ 9(\text{cos } x - \frac{2}{3})(\text{cos } x + \frac{1}{3}) \, we first distribute 9, then use algebraic simplifications to get our final expanded equation. This shows how algebraic manipulation transforms the problem into something manageable and solvable. Finally, matching terms allows us to determine the specific coefficients needed for B and C.