Problem 65 - Algebra and Trigonometry

Question

Use a double-or half-angle formula to solve the equation in the interval \([0,2 \pi) .\) $$\sin 2 x+\cos x=0$$

Step-by-Step Solution

Answer

Solutions are \( x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6} \).

Step 1: Identify Relevant Formulas

To solve the equation, identify that the double angle formula for sine can be used. The formula is \( \sin 2x = 2 \sin x \cos x \). Substitute this into the original equation.

Step 2: Substitute Double Angle Formula

Substitute \( \sin 2x = 2 \sin x \cos x \) into the equation, resulting in \( 2 \sin x \cos x + \cos x = 0 \).

Step 3: Factor the Equation

Factor out the common term \( \cos x \) from the equation: \( \cos x (2 \sin x + 1) = 0 \).

Step 4: Solve Each Factor

Set each factor to zero and solve: \( \cos x = 0 \) and \( 2 \sin x + 1 = 0 \).

Step 5: Solve \( \cos x = 0 \)

\( \cos x = 0 \) occurs at \( x = \frac{\pi}{2} \) and \( x = \frac{3\pi}{2} \) within the interval \([0, 2\pi)\).

Step 6: Solve \( 2 \sin x + 1 = 0 \)

Solve \( 2 \sin x + 1 = 0 \) by isolating \( \sin x \): \( \sin x = -\frac{1}{2} \). In the interval \([0, 2\pi)\), \( \sin x = -\frac{1}{2} \) occurs at \( x = \frac{7\pi}{6} \) and \( x = \frac{11\pi}{6} \).

Step 7: Compile Solutions

Collect all solutions from both factors, resulting in \( x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{7\pi}{6}, \text{and} \, \frac{11\pi}{6} \).

Key Concepts

Trigonometric Equations

Trigonometric equations are algebraic expressions that involve trigonometric functions like sine, cosine, and tangent. Solving these equations often requires the use of trigonometric identities or formulas to simplify and find solutions within a specific interval. For example, when faced with the equation \( \sin 2x + \cos x = 0 \), recognizing relevant identities such as the double angle formula for sine allows us to rewrite the equation into a more manageable form.

• Trigonometric equations can have multiple solutions due to the periodic nature of trigonometric functions.
• Key techniques include utilizing identities such as the Pythagorean, angle-sum, and double angle formulas to transform and simplify the equations.
• Ensure you understand the relationships between the various trigonometric functions to aid in simplification.

Substituting the double angle formula, \( \sin 2x = 2 \sin x \cos x \), into the equation transforms it into \( 2 \sin x \cos x + \cos x = 0 \). This restructuring facilitates the next steps in solving the equation.

Interval Solutions

Interval solutions refer to finding all the possible values for the variable within a specified range. In trigonometry, this is particularly important because trigonometric functions are periodic, repeating their values in regular intervals. For the interval \([0, 2\pi)\), you aim to find solutions that lie between 0 and \(2\pi\) without including \(2\pi\) itself.

• For \(\cos x = 0\), the solutions within \([0, 2\pi)\) are \(x = \frac{\pi}{2}\) and \(x = \frac{3\pi}{2}\).
• For \(2 \sin x + 1 = 0\), solving gives \(\sin x = -\frac{1}{2}\), and the solutions in the interval are \(x = \frac{7\pi}{6}\) and \(x = \frac{11\pi}{6}\).
• Always consider the basic period of the function when finding solutions and check if any transformations or shifts affect the results within the given interval.
• Write the solutions in terms of \(\pi\) when within a standard interval for easy understanding and communication.

By thoroughly exploring each factor within the interval, you can compile a full set of solutions, such as for this example: \(x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{7\pi}{6},\) and \(\frac{11\pi}{6}\).

Factoring Equations

Factoring is a powerful tool in solving trigonometric equations, particularly when an equation can be rewritten to extract common factors. In this problem, after substituting the double angle formula, the equation \(2 \sin x \cos x + \cos x = 0\) includes a common factor of \(\cos x\). By factoring, the equation transforms into \(\cos x (2 \sin x + 1) = 0\), enabling separate consideration of each factor.

• Factoring simplifies equations by breaking them into simpler components that can be solved individually.
• For trigonometric equations, look for algebraic common factors like \(\sin x, \cos x,\) or other multiples to make solving more direct.
• Once factored, set each component to zero to find potential solutions for the original equation.
• Always verify solutions to ensure none are extraneous due to the factoring process.

For example, solving \(\cos x = 0\) and \(2 \sin x + 1 = 0\) separately gives potential solutions, contributing to the complete set of solutions in the interval \([0, 2\pi)\). Factoring efficiently breaks down complex equations, making them easier to manage and solve. This method is particularly effective in reducing quadratic and higher-order trigonometric equations.