Chapter 6: Problem 14
Solve \(\sin 2 x=2 \cos x \cos 2 x\) algebraically. Give the general solution expressed in radians.
Short Answer
Expert verified
x = \frac{\pi}{6} + 2k\pi, \frac{5\pi}{6} + 2k\pi, \frac{3\pi}{2} + 2k\pi, where k \in \mathbb{Z}.
Step by step solution
01
- Use the Double Angle Formula
Rewrite \(\sin 2x\) using the double angle formula: \[\text{sin}(2x) = 2 \text{sin}(x) \text{cos}(x)\]
02
- Substitution
Substitute \(\text{sin}(2x)\) in the original equation with its equivalent expression: \[2 \text{sin}(x) \text{cos}(x) = 2 \text{cos}(x) \text{cos}(2x)\]
03
- Simplify and Cancel Out Common Terms
Divide both sides by \(2 \text{cos}(x)\), assuming \( \text{cos}(x) eq 0 \): \[ \text{sin}(x) = \text{cos}(2x) \]
04
- Use the Double Angle Formula for Cosine
Rewrite \( \text{cos}(2x) \) using the double angle formula: \[\text{cos}(2x) = 2 \text{cos}^2(x) - 1 \]
05
- Substitute and Reorganize
Replace \( \text{cos}(2x) \) in the equation: \[ \text{sin}(x) = 2 \text{cos}^2(x) - 1\] \ Rearrange to: \[ \text{sin}(x) + 1 = 2 \text{cos}^2(x) \]
06
- Use Pythagorean Identity
Use the identity \[ \text{sin}^2(x) + \text{cos}^2(x) = 1\] to replace \[ \text{cos}^2(x)\]: \[ \text{cos}^2(x) = 1 - \text{sin}^2(x) \] Substitute this into the equation: \[ \text{sin}(x) + 1 = 2(1 - \text{sin}^2(x))\]
07
- Solve the Quadratic Equation
Rearrange and solve the quadratic equation: \[ \text{sin}(x) + 1 = 2 - 2\text{sin}^2(x)\] \[ 2\text{sin}^2(x) + \text{sin}(x) - 1 = 0\] Use the quadratic formula \( ax^2 + bx + c = 0 \) to solve: \[ \text{sin}(x) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}\] \ Therefore, \[ \text{sin}(x) = \frac{1}{2}\] or \[ \text{sin}(x) = -1 \]
08
- Find the General Solution
For \[ \text{sin}(x) = \frac{1}{2}\], \[ x = \frac{\pi}{6} + 2k\pi \text{ or } x = \frac{5\pi}{6} + 2k\pi \text{ where } k \in \mathbb{Z} \text{(General solution for sine function)}\] \[ For \text{sin}(x) = -1, x = \frac{3\pi}{2} + 2k\pi \text{ where } k \in \mathbb{Z}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Double Angle Formulas
Double angle formulas are essential in trigonometry for expressing functions of double angles in terms of single angles. For sine and cosine, the formulas are:
\( \text{sin}(2\theta) = 2 \text{sin}(\theta) \text{cos}(\theta) \) and \( \text{cos}(2\theta) = \text{cos}^2(\theta) - \text{sin}^2(\theta) = 2 \text{cos}^2(\theta) - 1 = 1 - 2 \text{sin}^2(\theta) \).
These formulas help simplify expressions and solve equations involving double angles.
For example, in the equation \( \text{sin}(2x) = 2 \text{cos}(x) \text{cos}(2x) \), we use the formula \( \text{sin}(2x) = 2 \text{sin}(x) \text{cos}(x) \) to rewrite the equation. This step is crucial to make the equation more manageable.
By substituting and simplifying, we move towards solving the trigonometric equation.
\( \text{sin}(2\theta) = 2 \text{sin}(\theta) \text{cos}(\theta) \) and \( \text{cos}(2\theta) = \text{cos}^2(\theta) - \text{sin}^2(\theta) = 2 \text{cos}^2(\theta) - 1 = 1 - 2 \text{sin}^2(\theta) \).
These formulas help simplify expressions and solve equations involving double angles.
For example, in the equation \( \text{sin}(2x) = 2 \text{cos}(x) \text{cos}(2x) \), we use the formula \( \text{sin}(2x) = 2 \text{sin}(x) \text{cos}(x) \) to rewrite the equation. This step is crucial to make the equation more manageable.
By substituting and simplifying, we move towards solving the trigonometric equation.
Pythagorean Identity
The Pythagorean identity is a fundamental identity in trigonometry that relates the square of the sine and cosine functions:
\( \text{sin}^2(x) + \text{cos}^2(x) = 1 \).
This identity is derived from the Pythagorean theorem in a unit circle.
It is used to convert between \( \text{sin} \) and \( \text{cos} \) values.
In the provided solution, after simplifying the equation to \( \text{sin}(x) + 1 = 2 \text{cos}^2(x) \), we use the identity to replace \( \text{cos}^2(x) \) with \( 1 - \text{sin}^2(x) \):
\( \text{cos}^2(x) = 1 - \text{sin}^2(x) \).
Substituting this back into the equation, we help transform it into a quadratic equation in terms of \( \text{sin}(x) \).
Understanding and applying this identity is crucial in solving many trigonometric equations.
\( \text{sin}^2(x) + \text{cos}^2(x) = 1 \).
This identity is derived from the Pythagorean theorem in a unit circle.
It is used to convert between \( \text{sin} \) and \( \text{cos} \) values.
In the provided solution, after simplifying the equation to \( \text{sin}(x) + 1 = 2 \text{cos}^2(x) \), we use the identity to replace \( \text{cos}^2(x) \) with \( 1 - \text{sin}^2(x) \):
\( \text{cos}^2(x) = 1 - \text{sin}^2(x) \).
Substituting this back into the equation, we help transform it into a quadratic equation in terms of \( \text{sin}(x) \).
Understanding and applying this identity is crucial in solving many trigonometric equations.
Quadratic Equation
Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. These equations can be solved using the quadratic formula:
\( x = \frac{-b \, \pm \, \text{sqrt}(b^2 - 4ac)}{2a} \).
In the solution, we arrive at the quadratic equation \( 2\text{sin}^2(x) + \text{sin}(x) - 1 = 0 \).
To solve this, we identify \( a = 2 \), \( b = 1 \), and \( c = -1 \).
Plugging these into the quadratic formula, we solve for \( \text{sin}(x) \):
\( \text{sin}(x) = \frac{-1 \, \pm \, \text{sqrt}(1^2 - 4 \, \cdot 2 \, \cdot -1)}{2 \, \cdot 2} = \frac{-1 \, \pm \, 3}{4} \).
This provides us with two solutions: \( \text{sin}(x) = \frac{1}{2} \) and \( \text{sin}(x) = -1 \).
Understanding how to solve quadratic equations is key to solving many types of higher-order polynomial equations, including those involving trigonometric functions.
\( x = \frac{-b \, \pm \, \text{sqrt}(b^2 - 4ac)}{2a} \).
In the solution, we arrive at the quadratic equation \( 2\text{sin}^2(x) + \text{sin}(x) - 1 = 0 \).
To solve this, we identify \( a = 2 \), \( b = 1 \), and \( c = -1 \).
Plugging these into the quadratic formula, we solve for \( \text{sin}(x) \):
\( \text{sin}(x) = \frac{-1 \, \pm \, \text{sqrt}(1^2 - 4 \, \cdot 2 \, \cdot -1)}{2 \, \cdot 2} = \frac{-1 \, \pm \, 3}{4} \).
This provides us with two solutions: \( \text{sin}(x) = \frac{1}{2} \) and \( \text{sin}(x) = -1 \).
Understanding how to solve quadratic equations is key to solving many types of higher-order polynomial equations, including those involving trigonometric functions.
General Solutions of Trigonometric Functions
The general solution of trigonometric functions gives all possible values of a function. Because trigonometric functions are periodic, their solutions include a term for the period.
For instance, the general solution for \( \text{sin}(x) = \frac{1}{2} \) is:
\( x = \frac{\text{\pi}}{6} + 2k\text{\pi} \) or \( x = \frac{5\text{\pi}}{6} + 2k\text{\pi} \) where \( k \) is an integer (\( k \, \in \, \text{\mathbb{Z}} \)).
Similarly, for \( \text{sin}(x) = -1 \), the general solution is:
\( x = \frac{3\text{\pi}}{2} + 2k\text{\pi} \)
where \( k \, \in \, \text{\mathbb{Z}} \). These solutions represent all the angles for which the sine function has specific values over its period.
Being able to express general solutions is crucial as it encompasses all possible solutions for a trigonometric equation, not just a single one.
For instance, the general solution for \( \text{sin}(x) = \frac{1}{2} \) is:
\( x = \frac{\text{\pi}}{6} + 2k\text{\pi} \) or \( x = \frac{5\text{\pi}}{6} + 2k\text{\pi} \) where \( k \) is an integer (\( k \, \in \, \text{\mathbb{Z}} \)).
Similarly, for \( \text{sin}(x) = -1 \), the general solution is:
\( x = \frac{3\text{\pi}}{2} + 2k\text{\pi} \)
where \( k \, \in \, \text{\mathbb{Z}} \). These solutions represent all the angles for which the sine function has specific values over its period.
Being able to express general solutions is crucial as it encompasses all possible solutions for a trigonometric equation, not just a single one.