Problem 126

Question

Solve the equation.$$\cos x(\cos x+1)=0$$

Step-by-Step Solution

Verified

Answer

The solution to the equation $ \cos x(\cos x+1)=0 $ are $ x = \frac{\pi}{2} + k\pi $ and $ x = \pi + 2k\pi $, where $k$ is any integer.

1Step 1: Identify Factors

Looking at the equation $ \cos x(\cos x+1)=0 $, there are two factors in the equation: $ \cos x $ and $ \cos x + 1 $.

2Step 2: Set Factors to Zero

Set the factors to zero individually, $ \cos x = 0 $ and $ \cos x + 1 = 0 $. These are two separate equations that need to be solved.

3Step 3: Solve for the Variable

Solving for $x$, In the equation $ \cos x = 0 $, $x$ equals to $ \frac{\pi}{2} + k\pi $, where $k$ is an integer. In the equation $ \cos x + 1 = 0 $, $ \cos x = -1 $, thus, $x$ equals to $ \pi + 2k\pi $, where $k$ is an integer.

Key Concepts

cosine functionfactoring equationssolving equationsinteger solutions

cosine function

The cosine function, denoted as $ \cos x $, is a fundamental trigonometric function often encountered in many mathematical equations. It describes the horizontal component of an angle in a right triangle or a unit circle.
The cosine function has several distinct properties:

Periodicity: The cosine function is periodic with a period of $2\pi$. This means that its values repeat every $2\pi$ units.
Range: The range of the cosine function is between -1 and 1, inclusive. This means that $ \cos x $ can never be less than -1 or more than 1.
Even Function: Cosine is an even function, which means $ \cos(-x) = \cos x $.
Zeroes: The cosine function crosses zero at $x = \frac{\pi}{2} + k\pi$, where $k$ denotes integers. These points are crucial when determining when $ \cos x = 0 $.

Understanding these properties can significantly aid in solving equations involving the cosine function.

factoring equations

Factoring equations is an essential technique in algebra that involves expressing an equation as a product of its factors. This method is particularly useful in solving quadratic or polynomial equations. In the problem $ \cos x(\cos x+1)=0 $, factoring helps to identify potential solutions by setting each factor equal to zero.
Here’s how factoring plays out:

The given equation $ \cos x(\cos x+1)=0 $ is already factored into $ \cos x $ and $ \cos x + 1 $.
Each factor represents a potential solution. The rule is, if a product of factors equals zero, then at least one of the factors must equal zero.

By factoring, we split the equation into two separate manageable equations: $ \cos x = 0 $ and $ \cos x + 1 = 0 $. These simpler equations can be solved individually to find the values of $x$. This method cuts down complexity and allows for systematic solutions to more intricate equations.

solving equations

To solve equations, particularly trigonometric ones, it often requires setting each part of the equation to zero and finding when these conditions hold true. For the original equation $ \cos x(\cos x+1)=0 $, we simplify it by solving two simpler equations derived through factoring.
Steps to solve the equation effectively:

Solve $ \cos x = 0 $: Determine the values of $x$ where the cosine function equals zero. From the unit circle, this occurs at $x = \frac{\pi}{2} + k\pi$.
Solve $ \cos x + 1 = 0 $: This converts to $ \cos x = -1 $. From the unit circle, this solution is at $x = \pi + 2k\pi$.

Both solutions feature $k$, an integer, indicating that the solutions are periodic, appearing at regular intervals along the trigonometric circle.

integer solutions

In trigonometric equations, finding integer solutions involves identifying all possible values of $x$ that satisfy the equation for integer values of $k$. This involves leveraging the periodic nature of trigonometric functions.
For the equation $ \cos x(\cos x+1)=0 $, we found solutions:

For $ \cos x = 0 $, the integer solutions are $x = \frac{\pi}{2} + k\pi$.
For $ \cos x = -1 $, the integer solutions are $x = \pi + 2k\pi$.

Here, $k$ is any integer. This is because the cosine function is cyclic, and these solutions repeat indefinitely. Understanding these integer solutions helps in understanding the complete set of outcomes where the original equation holds true over its infinite cycle.

Problem 126

Problem 127

Other exercises in this chapter

Problem 125

Simplify the expression. $$\left(6.5 \times 10^{-6}\right)\left(3.8 \times 10^{4}\right)$$

View solution

Problem 126

Use DeMoivre's Theorem to find the indicated power of the complex number. Write the result in standard form. $$[4(\cos 2.8+i \sin 2.8)]^{5}$$

View solution

Problem 127

Use DeMoivre's Theorem to verify the indicated root of the real number. $-\frac{1}{2}(1+\sqrt{3} i)$ is a sixth root of 1.

View solution

Problem 127

Solve the equation.$$\sin x(2 \sin x+\sqrt{2})=0$$

View solution