Solving Trigonometric Equations By Finding All Solutions

The Organic Chemistry Tutor
20 Oct 201711:49
EducationalLearning
32 Likes 10 Comments

TLDRThis educational video script explores trigonometric equations involving sine, cosine, and tangent functions. It demonstrates how to find the values of x for various equations, such as sine x = 1/2, cosine x = -√2/2, and more complex scenarios like 2sin(x) + √3 = 0. The script explains the use of reference angles and the importance of considering all quadrants to find comprehensive solutions. It also covers how to generalize solutions using n for sine and cosine with a period of 2π, and for tangent with a period of π. The script concludes with simplifying the solution for sin^2(x) - 1 = 0, emphasizing the significance of understanding periodicity and the unit circle in trigonometry.

Takeaways
  • 📚 The script explains how to find the values of x for which sine x equals one-half using the unit circle, identifying angles at pi/6 and 5pi/6.
  • 🔍 To find all solutions, the script suggests adding multiples of 2pi to the initial solutions, resulting in the general solution x = pi/6 + 2pin, where n is an integer.
  • 📐 The script discusses finding angles for which cosine x equals negative square root of 2/2, identifying angles at 3pi/4 and 5pi/4 in the unit circle.
  • 🔄 For the cosine equation, the general solution is x = 3pi/4 + 2pin and x = 5pi/4 + 2pin, where n is any integer, covering all quadrants where cosine is negative.
  • 📉 The script solves an equation involving sine and square root of three, restricting solutions to the range 0 to 2pi, resulting in x = 4pi/3 and 5pi/3.
  • 🤔 For tangent-related problems, the script provides a method to find angles for which tangent x equals negative square root of 3/3, identifying angles at 5pi/6 and 11pi/6.
  • 🔢 The general solution for tangent equations is given as x = 5pi/6 + pin and x = 11pi/6 + pin, where n is any integer, considering the period of tangent as pi.
  • 🔄 The script also solves an equation involving sine squared, resulting in sine x being equal to plus or minus square root of 2/2, covering all four quadrants.
  • 🧩 The final solution for the sine squared equation is x = pi/4 + pi/2n, where n is any integer, simplifying the expression to represent all possible angles.
  • 📝 The script emphasizes the importance of writing solutions in terms of an integer variable n to account for all possible angles, especially for trigonometric functions with periodicity.
Q & A
  • What is the value of x if sine x equals one-half?

    -The values of x are pi/6 and 5pi/6, which correspond to 30 degrees and 150 degrees respectively. These are the angles where sine is equal to one-half in the unit circle.

  • Why are both pi/6 and 5pi/6 solutions to the equation sine x = 1/2?

    -Both pi/6 and 5pi/6 are solutions because sine is positive in quadrants one and two. Since sine of 30 degrees (pi/6) is 1/2, and sine is also positive at 150 degrees (5pi/6), both angles satisfy the equation.

  • How can you find all possible solutions for the equation sine x = 1/2?

    -To find all solutions, you can add multiples of 2pi to the initial solutions (pi/6 and 5pi/6). The general solution is x = pi/6 + 2pi*n or x = 5pi/6 + 2pi*n, where n is any integer.

  • What are the angles where cosine x equals negative square root of 2 divided by 2?

    -The angles are 135 degrees (3pi/4) and 225 degrees (5pi/4). These angles are in quadrants 2 and 3 where cosine is negative, and the reference angle is 45 degrees.

  • How do you determine the general solution for the equation cosine x = -√2/2?

    -The general solution is x = 3pi/4 + 2pi*n and x = 5pi/4 + 2pi*n, where n is any integer. This accounts for all coterminal angles that satisfy the equation.

  • What is the range of solutions for the equation 2sine x + √3 = 0 when restricted to 0 to 2pi?

    -The solutions in the range of 0 to 2pi are x = 4pi/3 and x = 5pi/3. These angles are in the specified range and satisfy the equation.

  • How do you find the reference angle for the equation 3tangent x + √3 = 0?

    -The reference angle is 30 degrees, as tangent of 30 degrees (pi/6) is equal to √3/3, which is the value we get after solving the equation for tangent x.

  • What are the general solutions for the equation 3tangent x + √3 = 0?

    -The general solutions are x = 5pi/6 + pi*n and x = 11pi/6 + pi*n, where n is any integer. This accounts for the periodicity of the tangent function, which is pi.

  • What does the equation 2sine squared x - 1 = 0 represent?

    -The equation represents a situation where the square of the sine function is equal to 1/2. After simplification, it leads to sine x being equal to ±√2/2.

  • What are the angles where sine x equals ±√2/2?

    -The angles are 45 degrees (pi/4), 135 degrees (3pi/4), 225 degrees (5pi/4), and 315 degrees (7pi/4). These angles are in the quadrants where sine is positive or negative as required.

  • How can you express all solutions for the equation 2sine squared x - 1 = 0?

    -The general expression for all solutions is x = pi/4 + pi/2*n, where n is any integer. This accounts for the 90-degree difference between the angles where sine x equals ±√2/2.

Outlines
00:00
📚 Solving Trigonometric Equations with Unit Circle

This paragraph discusses solving trigonometric equations using the unit circle. It begins with the equation sine x = 1/2, identifying angles such as 30° (π/6) and 150° (5π/6) where sine equals 1/2. The concept of coterminal angles is introduced, explaining how adding multiples of 2π to these angles yields all possible solutions for x. The general solution is given as x = π/6 + 2πn and x = 5π/6 + 2πn, where n is any integer. The paragraph then moves on to find solutions for cosine x = -√2/2, identifying angles in quadrants 2 and 3 where cosine is negative. Reference angles of 45° (π/4) and 135° (3π/4) are used to find all solutions, expressed as x = 3π/4 + 2πn and x = 5π/4 + 2πn. The section concludes with an example of solving 2sin(x) + √3 = 0 within the range of 0 to 2π, leading to solutions of x = 4π/3 and x = 5π/3.

05:02
🔍 Advanced Trigonometric Equations and Their Solutions

The second paragraph delves into more complex trigonometric equations. It starts with the equation 3tan(x) + √3 = 0, aiming to find all solutions by isolating tan(x) and simplifying to tan(x) = -√3/3. Using the 30-60-90 triangle, the reference angle of 30° is identified, and the angles in quadrants 2 and 4 where tangent is negative are calculated as 150° (5π/6) and 330° (11π/6). The general solution for tangent equations is given as x = 5π/6 + πn and x = 11π/6 + πn, where n is any integer. The paragraph then tackles the equation 2sin^2(x) - 1 = 0, which simplifies to sin^2(x) = 1/2. By taking the square root of both sides, the solutions for sin(x) are ±√2/2, leading to angles in all four quadrants: 45° (π/4), 135° (3π/4), 225° (5π/4), and 315° (7π/4). The final expression for all solutions is x = π/4 + π/2n, where n is any integer, providing a comprehensive method to find all possible values of x.

10:03
📐 Understanding Trigonometric Identities and Their Applications

The final paragraph focuses on the trigonometric identity involving sine squared. Starting with the equation 2sin^2(x) - 1 = 0, the paragraph simplifies it to sin^2(x) = 1/2, and then takes the square root of both sides to find sin(x) = ±√2/2. The solutions are analyzed in all four quadrants, where sine is positive in quadrants 1 and 2, and negative in quadrants 3 and 4, leading to the angles 45° (π/4), 135° (3π/4), 225° (5π/4), and 315° (7π/4). To generalize the solution for all possible values of x, the paragraph suggests adding 2πn to each solution, recognizing that each solution differs by π/2. The final expression for all solutions is x = π/4 + π/2n, where n is any integer, offering a simplified way to express all solutions to the trigonometric equation.

Mindmap
Keywords
💡sine
The sine function is a fundamental trigonometric function that relates the ratio of the opposite side to the hypotenuse in a right-angled triangle. In the context of the video, sine is used to find angles whose sine values are given, such as sine x = 1/2, which corresponds to angles of 30 degrees or π/6 and 150 degrees or 5π/6. The sine function is central to the theme of solving trigonometric equations for angles.
💡unit circle
A unit circle is a circle with a radius of 1, centered at the origin of a coordinate system. It is used in trigonometry to define the sine, cosine, and tangent of angles. The video script refers to the unit circle to determine the sine values of specific angles, such as sine of 30 degrees or π/6 being equal to 1/2.
💡quadrants
In trigonometry, the coordinate plane is divided into four quadrants. Quadrants are important for determining the signs of trigonometric functions. The video mentions that sine is positive in quadrants one and two, which helps in finding the correct angles for the given sine values.
💡coterminal angles
Coterminal angles are angles that share the same terminal side. They differ by full rotations, or multiples of 2π. The script explains how to find all solutions to a trigonometric equation by adding multiples of 2π to the initial angle, which results in coterminal angles.
💡cosine
Cosine is another basic trigonometric function that represents the ratio of the adjacent side to the hypotenuse in a right-angled triangle. The video discusses finding angles for which the cosine value is negative, such as cosine x = -√2/2, which corresponds to angles of 135 degrees or 3π/4 and 225 degrees or 5π/4.
💡reference angle
A reference angle is the acute angle formed by the terminal side of an angle and the x-axis. It is used to determine the value of trigonometric functions for any angle based on their values at standard angles. The script uses reference angles to find the cosine and sine values for specific angles in different quadrants.
💡tangent
The tangent function is the ratio of the opposite side to the adjacent side in a right-angled triangle, or equivalently, the sine divided by the cosine. The video script explores finding angles for which the tangent value is given, such as tangent x = -√3/3, which corresponds to angles in quadrants where tangent is negative.
💡period
The period of a function is the length of the smallest interval over which the function's values repeat. For sine and cosine, the period is 2π, and for tangent, it is π. The script explains how to write general solutions for trigonometric equations by considering the periods of the functions involved.
💡rationalize
Rationalizing a denominator involves expressing a fraction in a form that does not have an irrational number in the denominator. In the script, rationalizing is used to simplify expressions like 1/√2 to √2/2, making it easier to identify the sine values.
💡solutions
In the context of the video, solutions refer to the angles that satisfy a given trigonometric equation. The script demonstrates how to find all possible solutions, including those that are coterminal or within a specified range, by using trigonometric identities and properties.
Highlights

Sine x equals one-half can be solved using the unit circle, where sine of 30 degrees or pi/6 is equal to one-half.

Sine is positive in quadrants one and two, so x can also be 5 pi/6.

All solutions for sine x = 1/2 can be found by adding 2 pi to pi/6 and 5 pi/6, resulting in 13 pi/6 and 17 pi/6 respectively.

The general equation for all solutions of sine x = 1/2 is x = pi/6 + 2 pi n, where n is any integer.

Cosine x equals negative square root of 2/2 can be solved by considering cosine is negative in quadrants 2 and 3.

Reference angle for cosine 45 is the square root of 2/2, leading to solutions of 135 degrees or 3 pi/4 and 225 degrees or 5 pi/4.

The general solution for cosine x = -sqrt(2)/2 is x = 3 pi/4 + 2 pi n and x = 5 pi/4 + 2 pi n, where n is any integer.

For the equation 2 sine x + sqrt(3) = 0, solutions are restricted from 0 to 2 pi.

Isolating sine x gives sine x = -sqrt(3)/2, and the reference angle is 60 degrees or pi/3.

Solutions to the equation are x = 4 pi/3 and x = 5 pi/3, which are within the specified range.

The equation 3 tangent x + sqrt(3) = 0 leads to tangent x = -sqrt(3)/3.

The reference angle for tangent x = -sqrt(3)/3 is 30 degrees or pi/6, with tangent being negative in quadrants two and four.

Solutions include x = 5 pi/6 and x = 11 pi/6, with the period of tangent being pi.

The general form for solutions is x = 5 pi/6 + pi n and x = 11 pi/6 + pi n, where n is any integer.

For the equation 2 sine squared x - 1 = 0, the solution involves taking the square root of both sides, resulting in sine x = ±sqrt(2)/2.

Sine x can be positive or negative, leading to solutions in all four quadrants: pi/4, 3 pi/4, 5 pi/4, and 7 pi/4.

A general expression for all solutions is x = pi/4 + pi/2 n, where n is any integer, simplifying the solution set.

Alternatively, solutions can be written as adding 2 pi n to each individual solution for mathematical correctness.

Transcripts
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