Question

$$ \text { Solve the equation } \cos x+\cos 3 x+\cos 5 x+\cos 7 x=0 $$

Short answer

Question: Solve the equation \( \cos x + \cos 3x + \cos 5x + \cos 7x = 0 \) for x. Answer: The solutions for x are given by: 1. \(x = \frac{(2n+1)\pi}{2}\), where n is an integer. 2. \(x = (n+0.5)\pi\), where n is an integer. 3. \(x = (2n+1)\frac{\pi}{4}\), where n is an integer.

Step-by-step solution

Apply the sum-to-product identity for the first two terms

Applying the sum-to-product identity to \(\cos x\) and \(\cos 3x\), we have: $$ \cos x + \cos 3x = 2\cos\left(\frac{x+3x}{2}\right) \cos\left(\frac{x-3x}{2}\right) \implies 2 \cos 2x \cos -x $$ Considering the properties of the cosine function, \(\cos(-x) = \cos(x)\), the above expression becomes: $$ 2\cos 2x \cos x $$

Apply the sum-to-product identity for the last two terms

Applying the sum-to-product identity to \(\cos 5x\) and \(\cos 7x\), we have: $$ \cos 5x + \cos 7x = 2\cos\left(\frac{5x+7x}{2}\right) \cos\left(\frac{5x-7x}{2}\right) \implies 2 \cos 6x \cos -x $$ Considering the properties of the cosine function, \(\cos(-x) = \cos(x)\), the above expression becomes: $$ 2\cos 6x \cos x $$

Rewrite the given equation using the results from Steps 1 and 2

Now, we can rewrite the given equation using the results from Steps 1 and 2: $$ 2\cos 2x \cos x + 2\cos 6x \cos x = 0 $$

Solve the equation obtained in Step 3

From the equation obtained in Step 3, we can factor out \(2\cos x\): $$ 2\cos x (\cos 2x + \cos 6x) = 0 $$ This equation is satisfied if either \(\cos x = 0\) or \(\cos 2x + \cos 6x = 0\). Case 1: \(\cos x = 0\) In this case, we have x as an odd multiple of \(\frac{\pi}{2}\), i.e., \(x = \frac{(2n+1)\pi}{2}\), where n is an integer. Case 2: \(\cos 2x + \cos 6x = 0\) Applying the sum-to-product identity to this equation, we have: $$ 2 \cos 4x \cos(-2x) = 0 \implies 2 \cos 4x \cos 2x = 0 $$ Either \(\cos 2x = 0\) or \(\cos 4x = 0\). In either case, we can find the corresponding x values. For \(\cos 2x = 0\): \(2x = (2n+1)\frac{\pi}{2}\), so \(x = (n+0.5)\pi\), where n is an integer. For \(\cos 4x = 0\): \(4x = (2n+1)\frac{\pi}{2}\), so \(x = (2n+1)\frac{\pi}{4}\), where n is an integer.

Final solution

The solutions to the given equation are given by these x values: 1. \(x = \frac{(2n+1)\pi}{2}\), where n is an integer. 2. \(x = (n+0.5)\pi\), where n is an integer. 3. \(x = (2n+1)\frac{\pi}{4}\), where n is an integer.

Key concepts

cosine function

The cosine function is one of the fundamental functions in trigonometry, which describes the relationship between the angles and the lengths of the sides of a right-angled triangle. The cosine of an angle is the ratio of the adjacent side to the hypotenuse. It is represented as \( \cos(\theta) \).

This periodic function oscillates between -1 and 1, with a period of \( 2\pi \) radians (360 degrees). Understanding the symmetry of the cosine function is crucial. For example, cosine is an even function, meaning it holds that \( \cos(-x) = \cos(x) \). This property is vital in simplifying expressions and equations involving cosines.

Cosine values repeat in a predictable cycle:

• \( \cos(0) = 1 \)
• \( \cos(\pi/2) = 0 \)
• \( \cos(\pi) = -1 \)
• \( \cos(3\pi/2) = 0 \)
• \( \cos(2\pi) = 1 \)

These repeat every \( 2\pi \), making graphs of the cosine function look like smooth waves. The properties of the cosine function are essential when solving trigonometric equations, like the one presented in the exercise.

sum-to-product identities

Sum-to-product identities are a set of trigonometric identities that allow us to express sums or differences of trigonometric functions as products. These identities are particularly useful when simplifying expressions, solving equations, or integrating trigonometric functions.

In the context of the given exercise, the sum-to-product identity is used to transform the sum of two cosine terms into a product, turning a seemingly complex equation into a more manageable form. For instance:
\[ \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
This method was used to simplify both partial solutions \( \cos x + \cos 3x \) and \( \cos 5x + \cos 7x \).

By employing these identities, you can break down an equation into factors that are easier to solve, particularly useful when working through iterative problems involving multiple cosine terms. This identity is especially helpful in the algebraic manipulation required to solve the given trigonometric equation.

factorization in trigonometry

Factorization involves writing an expression as a product of its factors. This technique, common in algebra, is also quite effective in trigonometry, particularly when dealing with complex equations involving multiple trigonometric functions.

In our exercise, once we applied the sum-to-product identities, the next logical step was to factor what was obtained into simpler elements. Factoring out common terms, like \( 2\cos x \), from the rewritten expression:
\[ 2\cos 2x \cos x + 2\cos 6x \cos x = 0 \]
leads to:

• \( 2\cos x (\cos 2x + \cos 6x) = 0 \)

This simplifies the problem by reducing it to checking cases where either of the factors equals zero.

Factorization is critical in solving trigonometric equations because it identifies potential roots (solutions) by setting each factor equal to zero. Such techniques streamline the process of finding all possible solutions.

solving trigonometric equations

Solving trigonometric equations requires understanding periodicity and trigonometric identities. The process often involves identity manipulation (like sum-to-product) and clever algebraic techniques (like factorization) to reduce complex expressions into simpler forms.

In this exercise, we were tasked with solving a trigonometric equation that initially appeared overwhelming. By simplifying with trigonometric identities and factorization, we identified that the equation holds true for both:

• \( \cos x = 0 \)
• \( \cos 2x + \cos 6x = 0 \)

For \( \cos x = 0 \), we found solutions as odd multiples of \( \frac{\pi}{2} \).

For \( \cos 2x + \cos 6x = 0 \), further application of sum-to-product identities gave potential solutions with differing multipliers of \( \pi \). The solution involves recognising cycles and patterns inherent in trigonometric functions.
Combining these techniques allowed us to derive the detailed solutions \( x = \frac{(2n+1)\pi}{2} \), \( x = (n+0.5)\pi \), and \( x = (2n+1)\frac{\pi}{4} \), where \( n \) is an integer. This highlights the power of combining trigonometric identities with algebraic methods in solving equations.