Problem 29 - 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. $$2 \sin \theta=2 \cos 2 \theta$$

Step-by-Step Solution

Answer

\(\theta = 30^\circ, 150^\circ, 270^\circ\) or \(\theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}\).

Step 1: Divide Both Sides by 2

First, divide both sides of the equation by 2 to simplify the equation. This results in \(\sin \theta = \cos 2\theta\).

Step 2: Use Trigonometric Identity

Use the cosine double angle identity: \( \cos 2\theta = 1 - 2\sin^2 \theta \). Substitute this into the equation, giving \( \sin \theta = 1 - 2\sin^2 \theta \).

Step 3: Transform to Quadratic Equation

Rearrange the equation to form a quadratic in \(\sin \theta\): \( 2\sin^2 \theta + \sin \theta - 1 = 0 \). This can be solved using the quadratic formula: \( \sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \(a = 2\), \(b = 1\), and \(c = -1\).

Step 4: Solve the Quadratic Equation

Use the quadratic formula: \( \sin \theta = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4} \). This gives solutions \( \sin \theta = \frac{1}{2} \) and \( \sin \theta = -1 \).

Step 5: Find Corresponding Angles

Find \(\theta\) for \( \sin \theta = \frac{1}{2} \), which gives \(\theta = 30^\circ\) and \(\theta = 150^\circ\) within the range \([0^\circ, 360^\circ)\). For \( \sin \theta = -1 \), \(\theta = 270^\circ\) is the only solution within the range.

Step 6: Convert Angles to Radians

Convert \(\theta = 30^\circ\), \(150^\circ\), and \(270^\circ\) to radians: \(\theta = \frac{\pi}{6}\), \(\theta = \frac{5\pi}{6}\), and \(\theta = \frac{3\pi}{2}\) respectively.

Key Concepts

Radians and Degrees

When solving trigonometric equations, understanding the relationship between radians and degrees is fundamental. Radians and degrees are both units for measuring angles, but they are used in different contexts. It's crucial to be able to convert between the two, as scientific calculations often use radians.
Degrees, such as 30°, 150°, and 270° in our solution, are intuitive, often used in everyday situations like navigation and geography. However, in higher mathematics, particularly calculus, radians are preferred. This is because of the simpler and more natural integration with the circle's properties, like the unit circle, where the circumference is 2π radians.

• Degrees to radians: Multiply by π/180.
• Radians to degrees: Multiply by 180/π.

So, if we have \( \theta = 30^{\circ} \, 150^{\circ}, \;\text{and} 270^{\circ}\), these equate to \( \theta = \frac{\pi}{6}, \, \frac{5\pi}{6}, \, \text{and} \, \frac{3\pi}{2} \) radians, respectively. Whenever working with trigonometric equations, ensure you're comfortable converting these units to find accurate solutions.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for any variable angle. They are particularly helpful in simplifying and solving equations. In the exercise solved here, we used the cosine double angle identity:
\[ \cos 2\theta = 1 - 2\sin^2 \theta \] This identity is one of the many helpful snippets of information that allows us to transform equations and find solutions in different forms.

• Pythagorean Identity: \( \sin^2\theta + \cos^2\theta = 1 \)
• Angle Sum and Difference Identities
• Double and Half Angle Identities

Using such identities effectively can help change complex expressions into more familiar or simpler forms, which in turn makes solving trigonometric equations much less daunting. The use of these identities are abundant in problems involving trigonometry, enabling us to manipulate expressions and discover relationships between angles.

Quadratic Trigonometric Equations

Quadratic trigonometric equations involve trigonometric functions and require a combination of algebraic and trigonometric techniques to solve. In this problem, after applying the trigonometric identities, we transformed the equation into a quadratic in terms of \(\sin \theta\):
\[ 2\sin^2 \theta + \sin \theta - 1 = 0 \] We then employed the quadratic formula to find the solutions for \( \sin \theta \).

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

It's exactly like solving regular quadratic equations, except the solutions are angles that must fit within the specified range (often between 0° to 360° or 0 to 2π radians).
When solving these equations, identifying all potential solutions is crucial. Each solution of the quadratic equation needs further examination to determine the corresponding angles that fit the context, ensuring they fall within the desired domain, and converting them appropriately between radians and degrees if necessary. This problem highlights both the critical thinking and systematic approaches needed to solve such trigonometric equations effectively.