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Chapter 5: Problem 114

In Exercises \(97-116,\) use the most appropriate method to solve each equation on the interval \([0,2 \pi) .\) Use exact values where possible or give approximate solutions correct to four decimal places. $$7 \cos x=4-2 \sin ^{2} x$$

Short Answer

Expert verified
The solutions to the equation \(7 \cos x = 4 - 2 \sin^{2} x\) in the interval \([0,2\pi)\) are \(x = \frac{\pi}{3}\) and \(x = \frac{5\pi}{3}\).

Step by step solution

01

Convert the equation

Use the Pythagorean identity to express everything in terms of \(\cos x\). Specifically, substitute \(\sin^2 x = 1 - \cos^2 x\) into the original equation to get: \(7\cos x = 4 - 2(1 - \cos^2 x)\)
02

Simplify the equation

Simplify \(7\cos x = 4 - 2(1 - \cos^2 x)\) to \(7\cos x = 2\cos^2 x - 2\) and then rearrange to a standard quadratic equation form: \(2\cos^2 x - 7\cos x - 2 = 0\).
03

Solve the quadratic equation

Factor the quadratic equation \(2\cos^2 x - 7\cos x - 2 = 0\) to get \((2\cos x - 1)(\cos x + 2) = 0\). Setting each factor equal to zero gives the solutions for \(\cos x\), which is \(\frac{1}{2}\) and \(-2\)
04

Find the solutions for x

\(\cos x = \frac{1}{2}\) gives the solutions \(x = \frac{\pi}{3}\) and \(x = \frac{5\pi}{3}\) within the interval \([0,2\pi)\). The other solution \(\cos x = -2\) has no solutions since the value of cosine is always between -1 and 1
05

Check the solutions

Substitute the found solutions for \(x\) into the original equation to ensure that they are indeed correct.

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Key Concepts

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

Pythagorean Identity
The Pythagorean identity is a fundamental concept in trigonometry, and it's all about the relationship between sine and cosine. In its most basic form, it states that \( \sin^2 x + \cos^2 x = 1 \). This identity stems from the Pythagorean theorem, which applies to right-angled triangles.
When solving trigonometric equations, this identity can be handy for expressing one trigonometric function in terms of another.
  • If you know \( \sin x \), you can find \( \cos x \) using \( \cos^2 x = 1 - \sin^2 x \).
  • Conversely, knowing \( \cos x \), you can determine \( \sin x \).
For our problem, we use the identity to replace \( \sin^2 x \) with \( 1 - \cos^2 x \), allowing us to express everything in terms of \( \cos x \) and solve the equation more easily.
Quadratic Equation
Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \) where \( a \), \( b \), and \( c \) are constants. Solving these equations is a critical skill because they appear frequently in a variety of contexts.
There are several methods to solve quadratic equations, such as:
  • Factoring, which we use in this specific problem.
  • Using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
  • Completing the square.
In our exercise, transforming the trigonometric equation into a quadratic one in \( \cos x \) streamlines the problem, as it becomes easier to solve. Here, it took the form \( 2\cos^2 x - 7\cos x - 2 = 0 \), making it ready for factoring.
Cosine Function
The cosine function \( \cos x \) is one of the primary trigonometric functions. It relates an angle of a right triangle to the ratio of the length of the adjacent side over the hypotenuse.
Some crucial properties of the cosine function include:
  • Its range is from \(-1\) to \(1\).
  • It is periodic with a period of \(2\pi\).
  • It鈥檚 an even function, meaning \( \cos(-x) = \cos x \).
In our problem, knowing that the cosine must lie within \(-1\) and \(1\) helped us dismiss \(\cos x = -2\) as a solution, since this value falls outside of cosine's range. The solutions for \(\cos x\) were thus refined to obtain valid values.
Factoring
Factoring is a powerful algebraic technique used to solve equations, especially quadratics. Factoring involves writing an expression as a product of its factors, which are simpler expressions. In the context of quadratic equations, factoring can break down the equation into linear factors.
To factor successfully:
  • Look for common factors.
  • Recognize patterns such as differences of squares or perfect square trinomials.
In our exercise, we transformed \(2\cos^2 x - 7\cos x - 2 = 0\) into the product \((2\cos x - 1)(\cos x + 2) = 0\). This step simplified finding \(\cos x\) since each factor sets up a simple equation like \(2\cos x - 1 = 0\) or \(\cos x + 2 = 0\) that can be solved directly.

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Most popular questions from this chapter

In Exercises \(97-116,\) use the most appropriate method to solve each equation on the interval \([0,2 \pi) .\) Use exact values where possible or give approximate solutions correct to four decimal places. $$\cos ^{2} x+5 \cos x-1=0$$

In Exercises \(160-162,\) solve each equation on the interval \([0,2 \pi)\) Do not use a calculator. \(160.2 \mathrm{cos}\) $$\sin x+2 \sin \frac{x}{2}=\cos \frac{x}{2}+1$$

In Exercises \(85-96,\) use a calculator to solve each equation, correct to four decimal places, on the interval \([0,2 \pi)\) $$\cos ^{2} x-\cos x-1=0$$

In Exercises \(121-126,\) solve each equation on the interval \([0,2 \pi)\) \(2 \sin ^{3} x-\sin ^{2} x-2 \sin x+1=0\) (Hint: Use factoring by grouping.)

In Exercises \(97-116,\) use the most appropriate method to solve each equation on the interval \([0,2 \pi) .\) Use exact values where possible or give approximate solutions correct to four decimal places. $$5 \sec ^{2} x-10=0$$
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