Problem 54 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 39-62: Find the solutions of the equation that are in the interval \([0,2 \pi\) ). $$ \sqrt{3} \sin t+\cos t=1 $$

Step-by-Step Solution

Answer

The solutions are \( t = 0, \pi, \pi/3, 2\pi/3 \).

Step 1: Isolate one trigonometric function

Starting with the equation \( \sqrt{3} \sin t + \cos t = 1 \), let's isolate one of the trigonometric functions. We can express it as \( \cos t = 1 - \sqrt{3} \sin t \).

Step 2: Square both sides to eliminate square root

To remove the square root term, square both sides of the equation: \( (\cos t)^2 = (1 - \sqrt{3} \sin t)^2 \). This leads to \( \cos^2 t = 1 - 2\sqrt{3} \sin t + 3\sin^2 t \).

Step 3: Use Pythagorean identity

Recall the Pythagorean identity: \( \cos^2 t + \sin^2 t = 1 \). Substitute \( \cos^2 t \) in Step 2: \( 1 - \sin^2 t = 1 - 2\sqrt{3} \sin t + 3\sin^2 t \).

Step 4: Simplify the equation

Rearrange and simplify the equation: \( 0 = 2\sin^2 t - 2\sqrt{3} \sin t \), which simplifies to \( 2\sin^2 t - 2\sqrt{3} \sin t = 0 \).

Step 5: Factor the equation

Factor out \( 2\sin t \) from the equation: \( 2\sin t (\sin t - \sqrt{3}) = 0 \).

Step 6: Find solutions for each factor

Solve the two equations separately: 1) \( \sin t = 0 \) and 2) \( \sin t = \sqrt{3}/2 \).1) \( \sin t = 0 \) gives \( t = 0, \pi \).2) \( \sin t = \sqrt{3}/2 \) gives \( t = \pi/3, 2\pi/3 \).

Step 7: Verify solutions within the interval

Ensure that all solutions found are within the interval \([0, 2\pi)\). The solutions \( t = 0, \pi, \pi/3, and 2\pi/3 \) are all in the interval.

Key Concepts

Pythagorean Identity

The Pythagorean Identity is one of the fundamental relationships in trigonometry. It states that for any angle \( t \), the square of the sine plus the square of the cosine equals one: \[ \sin^2 t + \cos^2 t = 1 \]. This identity is derived from the Pythagorean Theorem applied to the unit circle, where the hypotenuse is the radius of the circle (always equal to 1). Consequently, when you draw a right triangle inside the unit circle, the horizontal leg corresponds to \( \cos t \) and the vertical leg to \( \sin t \). Both together, satisfying the identity.

• The identity helps simplify equations involving trigonometric expressions by allowing substitution.
• It plays a crucial role in verifying the truth of other trigonometric equations and identities.

In the context of solving trigonometric equations, as seen in the original exercise, substituting using the Pythagorean Identity can reduce the complexity of expressions, leading to easier factorization and solution finding.

Interval Notation

Interval notation is a way of describing a range of values and is particularly useful in defining domains for functions or solutions to equations. It helps to specify which solutions are valid within a given context.
For example, the interval \([0, 2\pi)\) indicates that our solutions should range from 0 to \(2\pi\), including 0 but excluding \(2\pi\). This is crucial when dealing with periodic functions like sine and cosine because they repeat every \(2\pi\) radians.

• Brackets like \([ \) and \( ] \) indicate inclusion of the endpoint.
• Parentheses \(( \) and \() \) indicate exclusion of the endpoint.

This approach ensures that we specify a clear and precise set of valid solutions, which is a key step in the problem-solving process. By checking all possible solutions against the interval, we can confidently determine which are correct and applicable.

Trigonometric Functions

Trigonometric functions are essential in understanding the relations between angles and distances in the unit circle. In trigonometry, the most common functions include sine (\(\sin\)), cosine (\(\cos\)), and tangent (\(\tan\)). These functions are periodic and have well-defined values for standard angles.

• The sine function \(\sin t\) measures the vertical component of an angle \(t\) in the unit circle.
• The cosine function \(\cos t\) measures the horizontal component.

In any trigonometric equation, understanding these functions and their properties is crucial. They are used to transform, manipulate, and solve expressions effectively.
In the example problem, solving for \(\sin t\) and \(\cos t\) involves isolating each function to determine their respective values, leading to the identification of potential solutions for the angle \(t\). This process is simplified by recognizing well-known angle values such as \(0, \pi, \pi/3,\) and \(2\pi/3\), which are frequently encountered in trigonometry.

Solution Verification

Solution verification involves checking each potential solution against the original constraints and ensuring they satisfy the given equation entirely. This step is critical, especially when an equation is modified during solving, for instance by squaring both sides. Such modifications can introduce "extraneous solutions"—solutions that solve the manipulated equation but not the original one.
Here's how you verify solutions:

• Check each solution within the specified interval.
• Substitute each solution back into the original equation to ensure it holds true.

In our exercise, after finding the potential solutions \(t = 0, \pi, \pi/3, \) and \(2\pi/3\), they are tested within the interval \([0, 2\pi)\) and confirmed to satisfy the original equation \(\sqrt{3} \sin t + \cos t = 1\). This final check assures that the solutions are correct and relevant to the problem's conditions, thus efficiently validating the problem-solving process.