Problem 37 - A Graphical Approach to Precalculus with Limits

Question

Solve each equation ( \(x\) in radians and \(\theta\) in degrees) for all exact solutions where appropriate. Round approximale values in radians to four decimal places and approximate values in degrees to the nearest tenth. Write answers using the least possible nonnegative angle measures. $$2 \cos ^{2} 2 \theta=1-\cos 2 \theta$$

Step-by-Step Solution

Answer

\( \theta = 30° + 180k \), \, 150° + 180k, \, 90° + 180k \) for integer \( k \).

Step 1: Simplify the equation

Start by simplifying the given equation: \[ 2\cos^{2}(2\theta) = 1 - \cos(2\theta) \] Let us make a substitution to simplify the equation further. Let \( x = \cos(2\theta) \). This gives us: \[ 2x^2 = 1 - x \]

Step 2: Rearrange the equation

Rearrange the equation to form a quadratic equation: \[ 2x^2 + x - 1 = 0 \] This is a standard quadratic equation of the form \( ax^2 + bx + c = 0 \).

Step 3: Solve the quadratic equation

Use the quadratic formula to solve for \( x \):\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For our equation, \( a = 2 \), \( b = 1 \), and \( c = -1 \). Plug these values into the quadratic formula: \[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{4} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4} \] This gives us two solutions: \( x = \frac{1}{2} \) and \( x = -1 \).

Step 4: Find the angles

Now, recall that \( x = \cos(2\theta) \). This means we need to solve: 1. \( \cos(2\theta) = \frac{1}{2} \) 2. \( \cos(2\theta) = -1 \)For \( \cos(2\theta) = \frac{1}{2} \), the angles are \(2\theta = \frac{\pi}{3} + 2k\pi\) and \(2\theta = \frac{5\pi}{3} + 2k\pi\).For \( \cos(2\theta) = -1 \), the angle is \(2\theta = \pi + 2k\pi\).

Step 5: Solve for θ

Solve for \( \theta \) from \( 2\theta \) using the angles found:1. From \( 2\theta = \frac{\pi}{3} + 2k\pi \), we get \( \theta = \frac{\pi}{6} + k\pi \).2. From \( 2\theta = \frac{5\pi}{3} + 2k\pi \), we get \( \theta = \frac{5\pi}{6} + k\pi \).3. From \( 2\theta = \pi + 2k\pi \), we get \( \theta = \frac{\pi}{2} + k\pi \).Thus, the solutions are all angles formed by: \( \theta = \frac{\pi}{6} + k\pi \), \( \frac{5\pi}{6} + k\pi \), \( \frac{\pi}{2} + k\pi \) for integer \( k \).

Step 6: Convert solutions to degrees

Convert the solutions from radians to degrees for easy understanding:1. \( \theta = \frac{\pi}{6} + k\pi \) becomes \( \theta = 30° + 180k \).2. \( \theta = \frac{5\pi}{6} + k\pi \) becomes \( \theta = 150° + 180k \).3. \( \theta = \frac{\pi}{2} + k\pi \) becomes \( \theta = 90° + 180k \).These are the angles in degrees, rounded to the nearest tenth when required.

Key Concepts

Radians and Degrees

In trigonometry, angles can be measured in two primary units: radians and degrees. Understanding the conversion between these two units is crucial for solving trigonometric equations. A full circle is 360 degrees; in radian measure, it is equivalent to \(2\pi\) radians. Thus, \(\pi\) radians makes up half of a circle or 180 degrees.

For conversion:

• To convert from degrees to radians, use the formula \(\text{radians} = \text{degrees} \times \frac{\pi}{180}\)
• Conversely, to convert from radians to degrees, use the formula \(\text{degrees} = \text{radians} \times \frac{180}{\pi}\)

In this exercise, we are asked to express solutions in both radians and degrees, rounding where necessary. This is important because trigonometric functions on a calculator usually accept arguments in radians.

Quadratic Equation

A quadratic equation is a second-order polynomial equation in a single variable, typically presented in the form \(ax^2 + bx + c = 0\). It is called 'quadratic' because it deals with the "square" of the variable, \(x^2\). Here, the solution involves using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to determine the roots of the given equation.

During our solution of the trigonometric equation, we transformed it into a quadratic by substituting \(x = \cos(2\theta)\), which simplified the equation to \(2x^2 + x - 1 = 0\). The steps include:

• Identifying coefficients \(a = 2\), \(b = 1\), \(c = -1\)
• Substituting these into the quadratic formula
• Solving for \(x\), which represents \(\cos(2\theta)\) in our context.

This approach allows us to handle the cosine terms algebraically before returning to the trigonometric functions they represent.

Cosine Function Solutions

When solving trigonometric equations, particularly those involving the cosine function, it's key to understand how its values can determine angles. If you have an equation like \(\cos(\theta) = \text{value}\), it is vital to recognize the general solutions based on common cosine values and their symmetries.

In our case:

• The equation \(\cos(2\theta) = \frac{1}{2}\) gives angles \(2\theta = \frac{\pi}{3}\) and \(2\theta = \frac{5\pi}{3}\). These reflect positions on the unit circle where the cosine value is \(\frac{1}{2}\).
• For \(\cos(2\theta) = -1\), the angle is \(2\theta = \pi\), indicative of the symmetry of cosine at these positions.

We then backtrack to find \(\theta\) as half of \(2\theta\). The end result is three sets of solutions for \(\theta\), each incorporating an integer \(k\) for their periodic nature, representing the repeating cycle of cosine across the unit circle.