Problem 35
Question
Exer. 1-38: Find all solutions of the equation. $$ \sin 2 x(\csc 2 x-2)=0 $$
Step-by-Step Solution
Verified Answer
The solutions are \( x = \frac{n\pi}{2} \), \( x = \frac{\pi}{12} + k\pi \), \( x = \frac{5\pi}{12} + k\pi \).
1Step 1: Understand the Equation
The given equation is \( \sin 2x (\csc 2x - 2) = 0 \). To solve it, we identify that if a product is zero, then at least one of the factors must be zero. Therefore, we need to solve the two separate equations: \( \sin 2x = 0 \) and \( \csc 2x - 2 = 0 \).
2Step 2: Solve \(\sin 2x = 0\)
The equation \( \sin 2x = 0 \) implies that \( 2x = n\pi \), where \( n \) is an integer. This results in \( x = \frac{n\pi}{2} \). Thus, the solutions from this part are \( x = \frac{n\pi}{2} \).
3Step 3: Solve \(\csc 2x - 2 = 0\)
Rewriting \( \csc 2x = 2 \) gives \( \sin 2x = \frac{1}{2} \). The solutions for \( \sin 2x = \frac{1}{2} \) are \( 2x = \frac{\pi}{6} + 2k\pi \) and \( 2x = \frac{5\pi}{6} + 2k\pi \), for integer \( k \). This results in solutions \( x = \frac{\pi}{12} + k\pi \) and \( x = \frac{5\pi}{12} + k\pi \).
4Step 4: Combine all solutions
Combining solutions, we have that \( x = \frac{n\pi}{2} \), \( x = \frac{\pi}{12} + k\pi \), and \( x = \frac{5\pi}{12} + k\pi \) for integers \( n \) and \( k \).
Key Concepts
Sine FunctionCosecant FunctionZero Product PropertyTrigonometric Identities
Sine Function
The sine function is a fundamental concept in trigonometry, represented as sin(x). It is a periodic function, meaning it repeats its values in regular intervals. The period of the sine function is \(2\pi\), and it oscillates between -1 and 1.
This function is particularly important in solving trigonometric equations, like in the exercise above. When dealing with the sine function, you often find yourself dealing with finding angles that satisfy certain properties.
One critical aspect is understanding when \(\sin(2x) = 0\). This equation holds true when \(2x = n\pi\), where \(n\) is any integer, because the sine of any integer multiple of \(\pi\) is zero.
You can think of it like the points where a wave crosses the horizontal axis on a graph. In simpler terms, the sine function helps define the wave's peak and valley points.
This function is particularly important in solving trigonometric equations, like in the exercise above. When dealing with the sine function, you often find yourself dealing with finding angles that satisfy certain properties.
One critical aspect is understanding when \(\sin(2x) = 0\). This equation holds true when \(2x = n\pi\), where \(n\) is any integer, because the sine of any integer multiple of \(\pi\) is zero.
You can think of it like the points where a wave crosses the horizontal axis on a graph. In simpler terms, the sine function helps define the wave's peak and valley points.
Cosecant Function
The cosecant function is the reciprocal of the sine function and is denoted as \(\csc(x)\). It is important to note that cosecant is undefined wherever sine equals zero, due to its reciprocal relationship.
In equations like \(\csc(2x) - 2 = 0\), we explore where the cosecant of double angles equals a specific value, in this case, 2. A pivotal step is to find the equivalent sine equations, since \(\csc(x) = \frac{1}{\sin(x)}\).
By setting \(\csc(2x) = 2\), we rearrange it to find \(\sin(2x) = \frac{1}{2}\). Solving this involves identifying those standard angles where the sine reaches \(\frac{1}{2}\), specifically \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\).
This shows that even seemingly complex trigonometric identities can be broken down into simpler trigonometric functions.
In equations like \(\csc(2x) - 2 = 0\), we explore where the cosecant of double angles equals a specific value, in this case, 2. A pivotal step is to find the equivalent sine equations, since \(\csc(x) = \frac{1}{\sin(x)}\).
By setting \(\csc(2x) = 2\), we rearrange it to find \(\sin(2x) = \frac{1}{2}\). Solving this involves identifying those standard angles where the sine reaches \(\frac{1}{2}\), specifically \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\).
This shows that even seemingly complex trigonometric identities can be broken down into simpler trigonometric functions.
Zero Product Property
The zero product property is a simple yet powerful algebraic principle used extensively in solving equations. It states that if a product of multiple factors equals zero, then at least one of the factors must be zero.
This property is crucial in solving our given equation: \(\sin 2x (\csc 2x - 2) = 0\). Applying this concept requires us to break the equation into two simpler ones: \(\sin 2x = 0\) and \(\csc 2x - 2 = 0\).
Both need to be solved separately to find all possible solutions of the original equation. This property simplifies complex multiplications to basic equations, offering an easy path to identify all solutions.
This property is crucial in solving our given equation: \(\sin 2x (\csc 2x - 2) = 0\). Applying this concept requires us to break the equation into two simpler ones: \(\sin 2x = 0\) and \(\csc 2x - 2 = 0\).
Both need to be solved separately to find all possible solutions of the original equation. This property simplifies complex multiplications to basic equations, offering an easy path to identify all solutions.
Trigonometric Identities
Trigonometric identities are mathematical expressions that relate different trigonometric functions to one another. They are essential tools in simplifying and solving trigonometric equations.
For instance, understanding the relationship between the sine and cosine or their reciprocal pairs, as seen in the transition from \(\csc(2x) = 2\) to \(\sin(2x) = \frac{1}{2}\).
Identifying such identities can make solving equations more straightforward, providing insight into different angles and solutions.
For instance, understanding the relationship between the sine and cosine or their reciprocal pairs, as seen in the transition from \(\csc(2x) = 2\) to \(\sin(2x) = \frac{1}{2}\).
Identifying such identities can make solving equations more straightforward, providing insight into different angles and solutions.
- Pythagorean identity: \(\sin^2(x) + \cos^2(x) = 1\)
- Reciprocal identity: \(\csc(x) = \frac{1}{\sin(x)}\)
- Even-odd identities: \(\sin(-x) = -\sin(x)\), etc.
Other exercises in this chapter
Problem 34
Exer. 25-36: Verify the reduction formula. $$ \tan (\pi-\theta)=-\tan \theta $$
View solution Problem 34
Express in terms of the cosine function with exponent \(1 .\) $$ \sin ^{4} \frac{\theta}{2} $$
View solution Problem 35
Exer. 1-50: Verify the identity. $$ \frac{\tan \alpha}{1+\sec \alpha}+\frac{1+\sec \alpha}{\tan \alpha}=2 \csc \alpha $$
View solution Problem 35
Exer. 25-36: Verify the reduction formula. $$ \tan \left(\theta+\frac{\pi}{2}\right)=-\cot \theta $$
View solution