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Chapter 8: Problem 4

Find all solutions of the equation. $$\sqrt{2} \cos x-1=0$$

Short Answer

Expert verified
Solutions are \(x = \frac{\pi}{4} + 2k\pi\) and \(x = \frac{7\pi}{4} + 2k\pi\), for any integer \(k\).

Step by step solution

01

Isolate the Trigonometric Function

To solve the equation \(\sqrt{2} \cos x - 1 = 0\), start by isolating the trigonometric function. Add 1 to both sides of the equation to get: \[ \sqrt{2} \cos x = 1. \] Then, divide both sides by \(\sqrt{2}\) to solve for \(\cos x\): \[ \cos x = \frac{1}{\sqrt{2}}. \] This equation can be rewritten as: \[ \cos x = \frac{\sqrt{2}}{2}. \]
02

Determine General Solutions

Recognize that \(\frac{\sqrt{2}}{2}\) is a common cosine value corresponding to standard angles. \(\cos x = \frac{\sqrt{2}}{2}\) means that \(x\) could be either \(\frac{\pi}{4}\) or \(\frac{7\pi}{4}\) within the range \([0, 2\pi]\). However, the cosine function is periodic with period \(2\pi\). Thus, the general solutions are: \[ x = \frac{\pi}{4} + 2k\pi \quad \text{and} \quad x = \frac{7\pi}{4} + 2k\pi, \] where \(k\) is any integer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cosine Function
The cosine function is a fundamental trigonometric function that relates to the coordinates of angles within a right-angled triangle. Specifically,
for an angle in a right-angled triangle, the cosine of the angle is the ratio of the length of the adjacent side to the length of the hypotenuse.
In terms of the unit circle, the cosine of an angle is the x-coordinate of the point on the circle corresponding to that angle.
  • Cosine values range from -1 to 1.
  • The cosine of 0 and multiples of r> are 1 or -1, as they lie directly on the x-axis, which represents the maximum or minimum values.
  • At a 90-degree angle (or r> ), the cosine value is 0, as it lies along the y-axis.
The cosine function is periodic and continuous, making it critical for modeling cyclical phenomena, like sound waves or light waves.
Periodic Functions
Periodic functions repeat their values in regular intervals. For trigonometric functions, this property is particularly noteworthy because it aids in predicting their behavior over an infinite range of inputs.
  • The cosine function is periodic with a period of r> . This means that r> will repeat its values every r> intervals.
  • A function's period is pivotal in determining the general solutions of trigonometric equations because it defines how often the solutions repeat.
  • For instance, any value of r> is equivalent to its r> counterpart, as seen when solving equations like r> , where solutions repeat every r> .
This periodic nature of trigonometric functions explains why we can extend solutions over an infinite domain by simply adding or subtracting integer multiples of the period, resulting in many solutions across different intervals.
General Solutions
Finding the general solutions of trigonometric equations involves identifying all possible angles that satisfy the given equation. In the case of r> we look for angles that return a specific cosine value, like r> .
The process for determining general solutions typically involves:
  • Identifying initial angles within a standard range (0 to r> ) that satisfy the equation. For example, r> and r> are both solutions for r> .
  • Understanding that due to the periodic nature of cosine, there are infinitely many solutions. This allows us to adjust the initial solutions by r> multiples, expressed as r> , where k is an integer.
  • Writing these angle expressions in a general format aids in finding all angles that solve the equation across the entire domain.
This approach results in a comprehensive representation of all solutions, making it straightforward to plug these into other problems or applications related to periodic behavior.

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Most popular questions from this chapter

Make the indicated trigonometric substitution in the given algebraic expression and simplify (see Example 7 ) Assume \(0 \leq \theta<\pi / 2\) $$ \frac{1}{x^{2} \sqrt{4+x^{2}}}, \quad x=2 \tan \theta $$

The displacement of a spring vibrating in damped harmonic motion is given by $$y=4 e^{-3 t} \sin 2 \pi t$$ Find the times when the spring is at its equilibrium position \((y=0)\)

Verify the identity. $$ (\tan x+\cot x)^{2}=\sec ^{2} x+\csc ^{2} x $$

Use a graphing device to find the solutions of the equation, correct to two decimal places. $$\frac{\cos x}{1+x^{2}}=x^{2}$$

The functions \(f(x)=\sin \left(\sin ^{-1} x\right) \quad\) and \(\quad g(x)=\sin ^{-1}(\sin x)\) both simplify to just \(x\) for suitable values of \(x\) . But these functions are not the same for all \(x\) . Graph both \(f\) and \(g\) to show how the functions differ. (Think carefully about the domain and range of \(\sin ^{-1} \).)
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