Problem 8 - Algebra and Trigonometry

Question

\(3-16 \cdot\) Solve the given equation. $$ 3 \sin 2 \theta-2 \sin \theta=0 $$

Step-by-Step Solution

Answer

The solutions for \( \theta \) are \( n\pi \) and \( \pm \cos^{-1}\left(\frac{1}{3}\right) + 2k\pi \).

Step 1: Identify the equation

We are given the equation: \( 3 \sin 2\theta - 2\sin \theta = 0 \). Our objective is to solve for \( \theta \).

Step 2: Use the Double Angle Identity

Recall the double angle identity: \( \sin 2\theta = 2\sin \theta \cos \theta \). Substitute this identity into the given equation: \[3(2\sin \theta \cos \theta) - 2\sin \theta = 0\]This simplifies to:\[6\sin \theta \cos \theta - 2\sin \theta = 0\]

Step 3: Factor the equation

Factor out \( \sin \theta \) from the equation:\[\sin \theta (6\cos \theta - 2) = 0\]This gives us two possible cases to solve.

Step 4: Solve \( \sin \theta = 0 \)

The first factor is \( \sin \theta = 0 \). Solutions occur when:\[\theta = n\pi\]where \( n \) is an integer.

Step 5: Solve \( 6\cos \theta - 2 = 0 \)

Set \( 6\cos \theta - 2 = 0 \) and solve for \( \theta \):\[6\cos \theta = 2\]\[\cos \theta = \frac{1}{3}\]The solutions are:\[\theta = \pm \cos^{-1}\left(\frac{1}{3}\right) + 2k\pi \]where \( k \) is an integer.

Step 6: Combine the solutions

Combine all the possible solutions for \( \theta \):\[\theta = n\pi, \pm \cos^{-1}\left(\frac{1}{3}\right) + 2k\pi\]where \( n \) and \( k \) are integers.

Key Concepts

Double Angle Identity

The double angle identity is a crucial concept in trigonometry, providing a way to simplify expressions involving angles. In this exercise, we use the identity for sine. Specifically, the double angle identity for sine is \( \sin 2\theta = 2 \sin \theta \cos \theta \).

This identity allows us to transform expressions involving \( \sin 2\theta \) into a product of sine and cosine functions. This transformation is particularly useful because it simplifies the process of solving trigonometric equations, as demonstrated in the solution.

By substituting \( \sin 2\theta \) with \( 2 \sin \theta \cos \theta \), we converted the original equation into a form that's easier to work with: \( 6\sin \theta \cos \theta - 2\sin \theta = 0 \).

Understanding and applying the double angle identities can greatly benefit when solving complex trigonometric equations or simplifying trigonometric expressions.

Factoring Equations

Factoring is a powerful algebraic method used to simplify and solve equations, including those involving trigonometric functions. After applying the double angle identity in our given equation, we arrived at \( 6\sin \theta \cos \theta - 2\sin \theta = 0 \).

At this point, factoring comes into play. Factoring involves finding common terms or expressions within an equation and simplifying it by grouping them. Here, we noticed that \( \sin \theta \) is a common factor in both terms, allowing us to factor it out as follows:

\( \sin \theta (6\cos \theta - 2) = 0 \).

This simplification splits the equation into two separate scenarios to solve:

• \( \sin \theta = 0 \)
• \( 6\cos \theta - 2 = 0 \)

Factoring helped break down the problem into smaller parts, making it easier to solve each component individually. Mastering factoring is essential for dealing with more complex algebraic and trigonometric equations efficiently.

Solutions of Trigonometric Functions

Finding solutions to trigonometric functions involves determining the values of the variable, often an angle, which satisfy the equation. In our exercise, after factoring the equation, we faced two scenarios to solve.

The first scenario was \( \sin \theta = 0 \). The solutions to this occur when \( \theta = n\pi \) where \( n \) is an integer. This is because the sine of \( \theta \) is zero at integer multiples of \( \pi \).

The second scenario involved solving \( 6\cos \theta - 2 = 0 \). Rearranging gives us \( \cos \theta = \frac{1}{3} \). The solutions can be found using the inverse cosine function leading to \( \theta = \pm \cos^{-1}\left(\frac{1}{3}\right) + 2k\pi \), where \( k \) is an integer. This expression accounts for the periodic nature of the cosine function.

Combining the solutions of both scenarios gives a complete set of possible values for \( \theta \). Understanding these solutions and the periodicity of trigonometric functions is vital for solving a range of problems in mathematics and applied sciences.