Problem 42 - Algebra 2

Question

Use identities to write each equation in terms of the single angle \(\theta .\) Then solve the equation for \(0 \leq \theta<2 \pi .\) $$ 4 \sin 2 \theta-3 \cos \theta=0 $$

Step-by-Step Solution

Answer

The solutions of the equation \(4 \sin 2 \theta-3 \cos \theta=0 \) within the interval \(0 \leq \theta<2 \pi \) are \( \theta = \pi / 2, 3\pi / 2, \arcsin(3/8), \pi - \arcsin(3/8)\).

Step 1: Replace \(\sin 2 \theta\)

Substitute \(\sin 2 \theta\) with \(2 \sin \theta \cos \theta\) using the double angle identity. The equation is now \(4(2 \sin \theta \cos \theta)-3 \cos \theta=0\), which simplifies to \(8 \sin \theta \cos \theta-3 \cos \theta=0\).

Step 2: Factorize

Factorize to group terms with \(\cos \theta\). The equation becomes \(\cos \theta (8 \sin \theta - 3) = 0\).

Step 3: Solve the Set of Equations

Consider \(8 \sin \theta - 3 = 0\) and \(\cos \theta = 0\) separately. From \(\cos \theta = 0\), we get \(\theta = \pi / 2, 3\pi / 2\). From \(8 \sin \theta - 3 = 0\), we get \(\sin \theta = 3/8\), \(\theta = \arcsin(3/8), \pi - \arcsin(3/8)\). Note that we need to consider the quadrants where \(\sin \theta\) is positive.

Step 4: Check the Validity of the Solution

Ensure that the solutions are within the given interval \(0 \leq \theta < 2\pi\). After verifying, we have 4 solutions \( \theta = \pi / 2, 3\pi / 2, \arcsin(3/8), \pi - \arcsin(3/8)\).

Key Concepts

Double Angle Identity

A key concept in trigonometry is the double angle identity for sine, which is expressed as: \[\sin 2\theta = 2 \sin \theta \cos \theta\]This identity is fundamental because it allows us to express trigonometric functions of double angles in terms of single angles, simplifying the solving process. When faced with the equation \(4 \sin 2 \theta - 3 \cos \theta = 0\), we substitute the double angle using the identity, giving us \(4(2 \sin \theta \cos \theta) - 3 \cos \theta = 0\). This simplifies to \(8 \sin \theta \cos \theta - 3 \cos \theta = 0\). Notice that replacing \(\sin 2\theta\) with its identity reduces the number of angle terms, making the problem easier to handle.

Trigonometric Equations

Solving trigonometric equations often requires strategically using various identities and manipulations to rewrite the equation in a simplified form. For this problem, after using the double angle identity, we are left with the equation \(8 \sin \theta \cos \theta - 3 \cos \theta = 0\). This equation is in a form where we can apply common algebraic methods such as factoring. By appropriately factoring the equation, we can separate it into simpler parts that are more straightforward to solve. This transformation allows us to recognize particular solutions and understand the relationships between the various trigonometric components involved in the equation.

Interval Solutions

When solving trigonometric equations, it's crucial to find solutions that exist within a given interval. The problem specifies the interval \(0 \leq \theta < 2\pi\), which requires the solutions to be within one complete cycle of a sine or cosine wave. In this context, the solutions to the equation \(\cos \theta = 0\) are \(\theta = \pi / 2\) and \(3\pi / 2\), both of which fit neatly within the interval. Additionally, solving \(8 \sin \theta - 3 = 0\) gives \(\sin \theta = 3/8\), leading to solutions \(\theta = \arcsin(3/8)\) and \(\pi - \arcsin(3/8)\). We confirm these solutions are within the designated interval, ensuring they are valid for the problem at hand.

Factoring Techniques

In solving the equation \(8 \sin \theta \cos \theta - 3 \cos \theta = 0\), factoring techniques play a crucial role. The term \(\cos \theta\) is common in both parts of the equation, so we factor it out, obtaining \(\cos \theta (8 \sin \theta - 3) = 0\).

• Factoring simplifies the equation by reducing it to the product of multiple expressions, making it easier to solve separately.
• This allows us to break down the complex equation into simpler, manageable parts: one where \(\cos \theta = 0\) and another where \(8 \sin \theta - 3 = 0\).
• After factoring, each part can be solved individually, and together they provide the complete set of solutions to the original equation.

By leveraging factoring, intricate trigonometric equations often become solvable using straightforward algebraic solutions.