Problem 111 - Algebra and Trigonometry

Question

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. $$ \cos ^{2} x+2 \cos x-2=0 $$

Step-by-Step Solution

Answer

The solutions to the given equation \(\cos ^{2} x+2 \cos x-2=0\) in the interval [0,2\pi) are \(x = \pi - \frac{\pi}{4}\) and \(x = \pi + \frac{\pi}{4}\)

Step 1: Rearrange the Equation

Rearrange the given equation \(\cos ^{2} x+2 \cos x-2=0\) in the form \(ax^{2} + bx + c = 0\). Here, the equation is already in that form, with \(\cos x\) being the variable.

Step 2: Factor the Equation

Factor the quadratic equation. This is always possible for real coefficients a, b and c with a ≠ 0.\n\nThe factored form is \((\cos x - \sqrt{2})(\cos x + \sqrt{2}) = 0\)

Step 3: Solve for cos x

Set each factor equal to zero and solve for \(\cos x\), which gives us the solutions \(\cos x = \sqrt{2}\) and \(\cos x = - \sqrt{2}\)

Step 4: Solve for x

Solve for x in the interval [0,2\pi). The equation \(\cos x = \sqrt{2}\) has no solutions in this interval as the maximum value of \(\cos x\) is 1. However, the equation \(\cos x = - \sqrt{2}\) gives us the solutions x = \(\pi \pm \frac{\pi}{4}\). Checking the given interval, both of these solutions fall within [0,2\pi) and are therefore valid.

Key Concepts

Understanding the Cosine Function

The cosine function, denoted as \(\cos x\), is a fundamental component of trigonometry, which helps relate the angles of a triangle to the lengths of its sides. This periodic function oscillates between -1 and 1, repeating every \(2\pi\) radians.

• At \(x = 0\) and \(x = 2\pi\), \(\cos x = 1\).
• At \(x = \pi\), \(\cos x = -1\).
• At critical points like \(x = \frac{\pi}{2}\) and \(x = \frac{3\pi}{2}\), \(\cos x = 0\).

A cosine function graph is a wave-like curve that starts at 1, decreases to -1 by \(\pi\), and returns to 1 at \(2\pi\). Applying the cosine function to solve trigonometric equations involves finding particular \(x\)-values where \(\cos x\) equals certain values.

Solving Quadratic Equations

When faced with a trigonometric equation such as \(\cos^2 x + 2\cos x - 2 = 0\), we can treat it similarly to a quadratic equation where \(\cos x\) is the variable. The general form of a quadratic equation is \(ax^2 + bx + c = 0\). Here, substituting \(y = \cos x\), we rewrite the equation as \(y^2 + 2y - 2 = 0\).
To solve it:

• Factor the equation when possible or use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• This step finds potential values for \(\cos x\).

For our example, factorizing gives \((\cos x - \sqrt{2})(\cos x + \sqrt{2}) = 0\), solving these factors provides candidate solutions for \(x\). Identifying factor pairs and quadratic solutions is crucial for proceeding to solve the trigonometric equation effectively.

Finding Solutions on an Interval

When working within a given interval, specifically \([0, 2\pi)\) in trigonometric contexts, we need to find all the \(x\)-values that fit within this range. These solutions are often considered within a closed interval from 0 to \(2\pi\), excluding \(2\pi\) itself, to avoid repetition due to the periodic nature of trigonometric functions.
- **Check Validity:** Once you solve the quadratic and find potential results for \(\cos x\), check which values are possible within the range of the cosine function. For example, \(\cos x = \sqrt{2}\) exceeds the maximum of 1 and thus isn't valid.- **Identify Suitable Solutions:** Values like \(\cos x = -\sqrt{2}\) are also impossible due to \(\cos x\) bounds but, hypothetically, if valid values existed, solve for \(x\) using inverse cosine functions.
Hence, determining valid solutions involves comparing possible solutions of \(\cos x\) with legitimate values within the \([-1,1]\) range and specific \(x\) intervals. These considerations are central to accurately solving trigonometric equations.