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Chapter 4: Problem 27

Solving Conditional Trigonometric Equations $$ 6 \sin ^{2}(x / 2)-5 \sin (x / 2)+1=0 $$

Short Answer

Expert verified
The solutions are \( x = \frac{\pi}{3} + 4k\pi \), \( x = \frac{5\pi}{3} + 4k\pi \), \( x = 2\arcsin(\frac{1}{3}) + 4k\pi \), and \( x = 2(\pi - \arcsin(\frac{1}{3})) + 4k\pi \).

Step by step solution

01

Identify the Substitution

First, let鈥檚 make a substitution to simplify. Let鈥檚 define a new variable: \( y = \sin(x/2) \). This changes the equation to: \( 6y^2 - 5y + 1 = 0 \).
02

Solve the Quadratic Equation

Now, solve the quadratic equation \( 6y^2 - 5y + 1 = 0 \). Use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 6 \), \( b = -5 \), and \( c = 1 \).
03

Substitute Values into Quadratic Formula

Substitute the values into the quadratic formula to get: \[ y = \frac{5 \pm \sqrt{25 - 24}}{12} \]. Simplify this to find: \[ y = \frac{5 \pm 1}{12} \]. This gives us two solutions: \( y = \frac{6}{12} = 0.5 \) and \( y = \frac{4}{12} = \frac{1}{3} \).
04

Substitute Back for \( \sin(x/2) \)

Recall that \( y = \sin(x/2) \). So we have: \( \sin(x/2) = 0.5 \) and \( \sin(x/2) = \frac{1}{3} \).
05

Solve for \( x \) from \( \sin(x/2) = 0.5 \)

Solve the equation \( \sin(x/2) = 0.5 \): The angle \( x/2 \) with \( \sin \) value 0.5 are \( \frac{\pi}{6} \) and \( \frac{5\pi}{6} \). Thus, \( x/2 = \frac{\pi}{6} + 2k\pi \) or \( x/2 = \frac{5\pi}{6} + 2k\pi \foreach integer k\). Solving for \( x \) we get: \( x = \frac{\pi}{3} + 4k\pi \) and \( x = \frac{5\pi}{3} + 4k\pi \).
06

Solve for \( x \) from \( \sin(x/2) = \frac{1}{3} \)

Solve the equation \( \sin(x/2) = \frac{1}{3} \): The angle \( x/2 \) with \( \sin \) value \( \frac{1}{3} \) are \( \arcsin(\frac{1}{3}) \) and \( \pi - \arcsin(\frac{1}{3}) \). Thus, \( x/2 = \arcsin(\frac{1}{3}) + 2k\pi \) or \( x/2 = \pi - \arcsin(\frac{1}{3}) + 2k\pi \). Solving for \( x \) we get: \( x = 2\arcsin(\frac{1}{3}) + 4k\pi \) and \( x = 2(\pi - \arcsin(\frac{1}{3})) + 4k\pi \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Substitution
In trigonometric equations, sometimes we encounter complex expressions that are difficult to solve directly. To simplify, we can use a technique called quadratic substitution. For instance, consider the equation: In this example, we substitute the trigonometric function with a new variable, making the equation easier to handle. Letting: This transforms the original equation into a quadratic form, which is easier to solve. Such substitutions help reduce the complexity of the trigonometric expressions, enabling us to focus on solving a simpler polynomial.
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that are true for all values of the variable. In solving trigonometric equations, it is often helpful to use these identities to simplify or transform equations into more solvable forms. For example, in our equation: Fields such as calculus, physics, and engineering utilize identities like:
  • It helps convert between different trigonometric functions and can reduce the complexity of equations. Recognizing and applying these identities can be highly beneficial for solving various problems in trigonometry and beyond.

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Most popular questions from this chapter

Solve each equation for the indicated variable. Solve \(x=2 \cos (3 y)\) for \(y\) where \(0 \leq y \leq \frac{\pi}{3}\)

Use a calculator to find the approximate value of each composition. Round answers to four decimal places. Some of these expressions are undefined. $$ \cot \left(\cos ^{-1}(-1 / \sqrt{7})\right) $$

Find the exact value of each composition without using a calculator or table. $$ \sec (\arcsin (1 / \sqrt{2})) $$

Find an equivalent algebraic expression for each composition. $$ \cos (\arctan (x)) $$

Solve each equation for the indicated variable. Solve \(q=3 \sin (\pi b-\pi)\) for \(b\) where \(\frac{1}{2} \leq b \leq \frac{3}{2}\).
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