Problem 67 - Precalculus with Limits

Question

In Exercises 63 - 74, use inverse functions where needed to find all solutions of the equation in the interval \( [0,2\pi) \). \( 2 \cos^2 x - 5 \cos x + 2 = 0 \)

Step-by-Step Solution

Answer

Based on the given solutions, the roots for \(x\) in the interval \( [0, 2\pi] \) are \(x = 0\) and \(x = 2\pi\).

Step 1 - Identifying the quadratic in terms of cos x

Rewrite the equation \(2 \cos^2 x - 5 \cos x + 2 = 0 \) as if it were a quadratic equation. Treat this equation as if it were in the form of \(ax^2 + bx + c = 0\), where \(a = 2\), \(b = -5\), and \(c = 2\). Here, the variable 'x' is replaced with \( \cos x \).

Step 2 - Calculating the roots using quadratic formula

The formula to find the roots of a quadratic equation is \((-b ± \sqrt{b^2 - 4ac}) / 2a\). Use this formula to find the roots which will give the solutions for \( \cos x \), which are \(x_1 = 1\) and \(x_2 = 2\).

Step 3 - Finding solution for x using arccosine

The next step is to find values of \(x\) that satisfy \( \cos x = 1\) and \( \cos x = 2\) within the given interval [0, 2π). Apply the arccosine function to solve for \(x\), also known as the inverse cosine function.

Step 4 - Evaluating for all possible x solutions within the given interval

Our range of values for \(x\) is [\(0, 2\pi\)]. Therefore, for \(\cos x = 1\), \(x = 0\) and \(x = 2\pi\) are solutions. For \(\cos x = 2\), there are no solutions, because the range of the cosine function is \([-1, 1]\), and 2 is outside this range. This means the equation \(2 \cos^2 x - 5 \cos x + 2 = 0\) has no solution in the interval [0, 2π) for \( \cos x = 2\).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are used to determine the angle associated with a given trigonometric value. These functions are essential in trigonometry for solving equations with trigonometric expressions. For example, where we need to find the value of \( x \) for a given \( \cos x \), we would use the arccosine, or inverse cosine function, denoted as \( \arccos \).

In the step by step solution provided for the exercise, we used the inverse cosine function to find the angle \( x \) for given cosine values. However, the range for arccosine is from \( 0 \) to \( \pi \), which aligns with the principal values of angles for the cosine function. Therefore, while looking for solutions within the interval \( [0, 2\pi) \), we need to remember that the inverse cosine will only provide us with the principal value, and we must consider the properties of cosine function to ascertain all possible solutions within the given interval.

Quadratic Formula

Solving quadratic equations is a fundamental part of algebra. A quadratic equation has the standard form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) represents an unknown variable. The quadratic formula \( (-b \pm \sqrt{b^2 - 4ac}) / 2a \) is the universal method for finding the roots of any quadratic equation.

In our exercise, we apply the quadratic formula to solve for \( \cos x \), treating it as the variable. Critical to this is recognizing that the equation \( 2 \cos^2 x - 5 \cos x + 2 = 0 \) is a quadratic in cosine, a powerful technique for solving trigonometric equations. But remember, when applying this formula, the discriminant (the part under the square root, \( b^2-4ac \)) must be non-negative for the equation to have real, and in turn, meaningful trigonometric solutions.

Cosine Function Interval

The cosine function is periodic and has a range of \( [-1, 1] \). This means that the values outputted by the cosine function will always be within this interval. The significance of this property becomes clear when solving trigonometric equations; any solution that falls outside this range is not valid.

In solving our exercise, the value \( \cos x = 2 \) was found. Since this is outside the cosine function's range, it does not provide a valid solution. Therefore, when solving trigonometric equations and obtaining a cosine value, always ensure it is within the acceptable range of \( -1 \) to \( 1 \), which helps in identifying the imaginable (feasible) solutions from the impossible (or unfeasible) ones.