Problem 74 - Contemporary Precalculus

Question

Use factoring, the quadratic formula, or identities to solve the equation. Find all solutions in the interval \([0,2 \pi)\). $$\cos ^{2} x+5 \cos x=1$$

Step-by-Step Solution

Answer

Question: Solve the equation \(\cos ^{2} x+5 \cos x=1\) for x in the interval \([0, 2\pi)\). Answer: The solutions are \(x_1 = \pi + \arccos{(\frac{-5 + \sqrt{29}}{2})}\) and \(x_2 = \pi - \arccos{(\frac{-5 + \sqrt{29}}{2})}\).

Step 1: Rewrite the equation in standard quadratic form

Write the equation as \(\cos ^{2} x+5 \cos x-1=0\).

Step 2: Apply the quadratic formula

The given equation is in the form \(ax^2 + bx + c = 0\) with \(a=1\), \(b=5\), and \(c=-1\). Applying the quadratic formula, we have: $$\cos x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Substitute \(a=1\), \(b=5\), and \(c=-1\) into the formula: $$\cos x = \frac{-5 \pm \sqrt{5^2 - 4(1)(-1)}}{2(1)}$$ Simplify: $$\cos x = \frac{-5 \pm \sqrt{29}}{2}$$

Step 3: Find the solutions for x

We have two possible values of \(\cos x\): $$\cos x_1 = \frac{-5 + \sqrt{29}}{2}$$ $$\cos x_2 = \frac{-5 - \sqrt{29}}{2}$$ To find the solutions for x, we need to find the inverse cosine (arccos) of both \(x_1\) and \(x_2\): $$x_1 = \arccos{(\frac{-5 + \sqrt{29}}{2})}$$ $$x_2 = \arccos{(\frac{-5 - \sqrt{29}}{2})}$$ Since the value of \(\cos x\) is negative for both solutions, we know the angle x must be in the second or third quadrant (\(\pi

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool used to solve quadratic equations, which are equations in the form of \(ax^2 + bx + c = 0\). It helps find the values of \(x\) that satisfy the equation, thus solving the problem. The formula is written as:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Here, \(a\), \(b\), and \(c\) are coefficients from the equation.
If the quadratic equation represents a trigonometric expression, replacing \(x\) with a trigonometric function like \(\cos x\) or \(\sin x\) is common. This allows us to solve for specific angles.
In our problem, we applied the quadratic formula to \(\cos x\), leading to potential solutions needing further evaluation using inverse trigonometric functions.

Factoring

Factoring is a method of solving quadratic equations by expressing the equation as a product of its factors. This method aims to set each factor equal to zero and solve for \(x\).
While factoring can sometimes give quick solutions, not all quadratic equations are easily factored, especially when coefficients result in complex numbers or complicated roots. For example:

• An equation like \(\cos^2 x + 5\cos x - 1 = 0\) can be difficult to factor directly into nice integers.

When factoring isn't feasible after attempting various common approaches (like grouping or using special products), methods such as the quadratic formula come into play. These methods allow for comprehensive solutions of any quadratic, ensuring no root goes missing.

Inverse Trigonometric Functions

Inverse trigonometric functions are essential tools for finding angles given their trigonometric values. When solving the equation \(\cos^2 x + 5\cos x - 1 = 0\), we found possible values for \(\cos x\) which are expressed as fractions. To determine the angle \(x\), we use the inverse cosine, denoted as \(\arccos\).

• These functions help us reverse the usual trigonometric process: instead of calculating the cosine of an angle, we are finding the angle that has a certain cosine value.
• It's crucial to consider the quadrant in which the angle lies, as angles generated by \(\arccos\) are typically in the range of \([0, \pi]\).

For negative cosine values, \(x\) resides in the second or third quadrants. In our case, angles \(x_1\) and \(x_2\) are calculated within intervals accounting for these conditions.

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Problem 74

Other exercises in this chapter

Problem 73

Show that the restricted cosecant function, whose domain consists of all numbers \(x\) such that \(-\pi / 2 \leq x \leq \pi / 2\) and \(x \neq 0,\) has an inver

View solution

Problem 73

Prove the identity. $$\frac{\cos x \cot x}{\cot x-\cos x}=\frac{\cot x+\cos x}{\cos x \cot x}$$

View solution

Problem 74

Show that the restricted cotangent function, whose domain is the interval \((0, \pi),\) has an inverse function. Sketch its graph.

View solution

Problem 74

Prove the identity. $$\frac{\cos ^{3} x-\sin ^{3} x}{\cos x-\sin x}=1+\sin x \cos x$$

View solution

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool used to solve quadratic equations, which are equations in the form of \(ax^2 + bx + c = 0\). It helps find the values of \(x\) that satisfy the equation, thus solving the problem. The formula is written as:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Here, \(a\), \(b\), and \(c\) are coefficients from the equation.
If the quadratic equation represents a trigonometric expression, replacing \(x\) with a trigonometric function like \(\cos x\) or \(\sin x\) is common. This allows us to solve for specific angles.
In our problem, we applied the quadratic formula to \(\cos x\), leading to potential solutions needing further evaluation using inverse trigonometric functions.

Factoring

Factoring is a method of solving quadratic equations by expressing the equation as a product of its factors. This method aims to set each factor equal to zero and solve for \(x\).
While factoring can sometimes give quick solutions, not all quadratic equations are easily factored, especially when coefficients result in complex numbers or complicated roots. For example:

• An equation like \(\cos^2 x + 5\cos x - 1 = 0\) can be difficult to factor directly into nice integers.

When factoring isn't feasible after attempting various common approaches (like grouping or using special products), methods such as the quadratic formula come into play. These methods allow for comprehensive solutions of any quadratic, ensuring no root goes missing.

Inverse Trigonometric Functions

Inverse trigonometric functions are essential tools for finding angles given their trigonometric values. When solving the equation \(\cos^2 x + 5\cos x - 1 = 0\), we found possible values for \(\cos x\) which are expressed as fractions. To determine the angle \(x\), we use the inverse cosine, denoted as \(\arccos\).

• These functions help us reverse the usual trigonometric process: instead of calculating the cosine of an angle, we are finding the angle that has a certain cosine value.
• It's crucial to consider the quadrant in which the angle lies, as angles generated by \(\arccos\) are typically in the range of \([0, \pi]\).

For negative cosine values, \(x\) resides in the second or third quadrants. In our case, angles \(x_1\) and \(x_2\) are calculated within intervals accounting for these conditions.