Problem 57 - Precalculus: Functions and Graphs

Question

Use inverse trigonometric functions to find the solutions of the equation that are in the given interval, and approximate the solutions to four decimal places. \(15 \cos ^{4} x-14 \cos ^{2} x+3=0\) \([0, \pi]\)

Step-by-Step Solution

Answer

The solutions in the interval [0, π] are approximately 0.5880, 2.5536, 0.9553, and 2.1863.

Step 1: Simplify the Equation

Notice that the equation is in terms of \(\cos^4 x\) and \(\cos^2 x\). To simplify this, we use a substitution: let \(u = \cos^2 x\). Therefore, the equation becomes a quadratic equation: \(15u^2 - 14u + 3 = 0\).

Step 2: Solve the Quadratic Equation

Applying the quadratic formula, \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 15\), \(b = -14\), and \(c = 3\), substitute these values into the formula. This yields: \[u = \frac{14 \pm \sqrt{(-14)^2 - 4 \cdot 15 \cdot 3}}{30}\] which simplifies to: \[u= \frac{14 \pm \sqrt{196 - 180}}{30} = \frac{14 \pm \sqrt{16}}{30} = \frac{14 \pm 4}{30}\]Hence, the solutions for \(u\) are \(u = \frac{18}{30} = \frac{3}{5}\) and \(u = \frac{10}{30} = \frac{1}{3}\).

Step 3: Convert Back from the Substitution

Since \(u = \cos^2 x\), we get two equations: 1. \(\cos^2 x = \frac{3}{5}\)2. \(\cos^2 x = \frac{1}{3}\).

Step 4: Solve for \(x\) Using Inverse Cosine

For \(\cos^2 x = \frac{3}{5}\), take the square root to find \(\cos x = \pm\sqrt{\frac{3}{5}}\).Thus, find \(x\) using the inverse cosine:- \(x = \cos^{-1}\left(\sqrt{\frac{3}{5}}\right)\)- \(x = \cos^{-1}\left(-\sqrt{\frac{3}{5}}\right)\)Similarly, for \(\cos^2 x = \frac{1}{3}\), - \(x = \cos^{-1}\left(\sqrt{\frac{1}{3}}\right)\)- \(x = \cos^{-1}\left(-\sqrt{\frac{1}{3}}\right)\).

Step 5: Find Solutions in the Interval \([0, \pi]\)

Calculate the principal values:1. \(x = \cos^{-1}\left(\sqrt{\frac{3}{5}}\right) \approx 0.5880\)2. \(x = \cos^{-1}\left(-\sqrt{\frac{3}{5}}\right) \approx 2.5536\)3. \(x = \cos^{-1}\left(\sqrt{\frac{1}{3}}\right) \approx 0.9553\)4. \(x = \cos^{-1}\left(-\sqrt{\frac{1}{3}}\right) \approx 2.1863\)All these solutions are within \([0, \pi]\).

Key Concepts

Quadratic Equations

Quadratic equations are polynomial equations of the second degree, typically written in the form \(ax^2 + bx + c = 0\). Here:

• \(a\), \(b\), and \(c\) are constants with \(a eq 0\).
• The solutions, also known as roots, can be found using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

When dealing with equations involving higher powers, like \(\cos^4 x\), we can often simplify them into quadratic form by substitution, as we did with \(u = \cos^2 x\). This method reduces the complexity, transforming it into a manageable quadratic equation. In our specific example, the quadratic equation \(15u^2 - 14u + 3 = 0\) was solved using the quadratic formula. Remember, solving quadratic equations is a fundamental skill, helping in various algebra and calculus problems.

Cosine Function

The cosine function, denoted as \(\cos(x)\), is one of the primary trigonometric functions. It relates to the adjacent side and hypotenuse of a right-angle triangle. The function is periodic with a period of \(2\pi\), meaning it repeats its values every \(2\pi\) units. In the interval \([0, \pi]\), the cosine function decreases from \(1\) to \(-1\). Here are some key points:

• The cosine function is even, meaning \(\cos(x) = \cos(-x)\).
• Its values range from \(-1\) to \(1\).
• \(\cos(x) = 0\) at \(x = \frac{\pi}{2}\).

In problems like ours, understanding the behavior of the cosine function helps us find solutions constrained to specific intervals. We look for angles within the given interval that satisfy the trigonometric equation. This step is crucial for deriving accurate principal solutions.

Trigonometric Equations

Trigonometric equations involve trigonometric functions such as sine, cosine, and tangent. These equations require solutions that often utilize identities and inverse trigonometric functions to find angles satisfying the given conditions. Some strategies for tackling trigonometric equations include:

• Substitution to convert complex expressions into simpler ones, as seen with \(\cos^2 x\).
• Using inverse functions to find specific angle measures.
• Simplifying using trigonometric identities when possible.

In our example, we formed a trigonometric equation by solving \(15 \cos^4 x - 14 \cos^2 x + 3 = 0\) through substitution that led to a quadratic equation. Afterward, inverse trigonometric functions were used to derive the angles corresponding to specific cosine values. Mastery of trigonometric equations allows solving diverse problems by considering possible solutions across one or more periods of the trigonometric function involved.

Principal Values

Principal values in trigonometry refer to the primary solutions of inverse trigonometric functions within a predetermined interval. For the cosine function, the principal value is typically found in the interval \([0, \pi]\). When using

• \(\cos^{-1}\) to solve an equation, the result gives the angle whose cosine equals the given value.
• In our example, we used \(\cos^{-1}\) to find angles corresponding to \(\cos x = \pm\sqrt{\frac{3}{5}}\) and \(\cos x = \pm\sqrt{\frac{1}{3}}\) within \([0, \pi]\).

Principal values provide critical solutions that are valid within specific bounds, ensuring they conform to expected constraints like those set by a particular interval. Understanding this concept is vital because it distinguishes primary results from possibly countless other solutions differing by whole periods of the periodic function.