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Magazine Chapter 20: Problem 11
Solve the given trigonometric equations analytically (using identities when necessary for exact values when possible) for values of \(x\) for \(0 \leq x<2 \pi\). $$2 \sin ^{2} x-\sin x=0$$
Short Answer
Expert verified
Solutions are \(x = 0, \pi, \frac{\pi}{6}, \frac{5\pi}{6}\).
Step by step solution
01
Factoring the Equation
The given equation is \(2 \sin^2 x - \sin x = 0\). Start by factoring out a common factor of \(\sin x\):\[ \sin x (2\sin x - 1) = 0 \] This equation is now easily separable into two individual equations.
02
Solving Each Factor for Zero
Set each factor equal to zero:1. \( \sin x = 0 \)2. \( 2\sin x - 1 = 0 \)
03
Solving \(\sin x = 0\)
For \( \sin x = 0 \), \(x\) can be any point where the sine is zero within the interval \([0, 2\pi)\). The solutions are:\[ x = 0, \pi \] because these are the standard angles within the given range where the sine function equals zero.
04
Solving \(2\sin x - 1 = 0\)
For \(2\sin x - 1 = 0\), solve for \(\sin x\):\[ 2\sin x = 1 \quad \Rightarrow \quad \sin x = \frac{1}{2} \]The sine function equals \(\frac{1}{2}\) at two points within \([0, 2\pi)\):\[ x = \frac{\pi}{6}, \frac{5\pi}{6} \] since these are the reference angles for which sine is positive.
05
Combining All Solutions
Combine all solutions obtained:\[ x = 0, \pi, \frac{\pi}{6}, \frac{5\pi}{6} \]These are all the values of \(x\) in the interval \([0, 2\pi)\) that satisfy the original equation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Factoring
When faced with a trigonometric equation like \(2 \sin^2 x - \sin x = 0\), factoring becomes a handy tool to simplify the equation. Factoring involves rewriting an expression as a product of its factors, which in this case means finding a common term in all parts of the equation.
In our example, \(\sin x\) is a common factor in both terms. By factoring it out, we reshape the equation into \( \sin x (2\sin x - 1) = 0 \). This transformation breaks it into smaller, more manageable pieces.
Here's how factoring simplifies solving equations:
In our example, \(\sin x\) is a common factor in both terms. By factoring it out, we reshape the equation into \( \sin x (2\sin x - 1) = 0 \). This transformation breaks it into smaller, more manageable pieces.
Here's how factoring simplifies solving equations:
- It converts a single complex equation into simpler parts.
- Each factor set to zero becomes a new, much easier equation to solve.
- We avoid complex algebra and stick to simple trigonometric terms.
Sine Function
The sine function, key to understanding trigonometric problems, maps angles to their sine values, ranging from -1 to 1. It is often depicted as \( \sin x \) and is periodic, repeating every \(2\pi\) radians.
For the equation component \(\sin x = 0\), we look for angles where the sine value is precisely zero. The critical points within one full cycle \([0, 2\pi)\) are 0 and \(\pi\), corresponding to the sine wave crossing the x-axis.
Understanding these points helps identify where certain sine values occur:
For the equation component \(\sin x = 0\), we look for angles where the sine value is precisely zero. The critical points within one full cycle \([0, 2\pi)\) are 0 and \(\pi\), corresponding to the sine wave crossing the x-axis.
Understanding these points helps identify where certain sine values occur:
- \(\sin x = 0\) at angles like 0, \(\pi\), and \(2\pi\) (out of range for our specific interval).
- \(\sin x = \frac{1}{2}\) implies moving away from these points, known as reference angles.
Unit Circle
The unit circle is an essential tool for trigonometry, especially solving equations involving sine and cosine. It is a circle with a radius of one, centered at the origin of the coordinate plane. Here's how it helps with our equation:
Knowing where certain sine values occur on the unit circle allows us to find the exact angles needed in trigonometric equations, such as:
Knowing where certain sine values occur on the unit circle allows us to find the exact angles needed in trigonometric equations, such as:
- \(\sin x = 0\) at angles where the terminal side of angle \(x\) intersects the x-axis, such as 0 and \(\pi\).
- For \(\sin x = \frac{1}{2}\), the known angles from the unit circle that match are \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\).
Solving Equations
Solving trigonometric equations involves isolating the trigonometric function and manipulating the equation to find all possible solutions within a specified interval.
Let's see how it applies:
Let's see how it applies:
- Begin by factoring any common terms to simplify the equation's structure.
- Set each factor to zero, creating solvable equations, like \(\sin x = 0\) or \(2\sin x - 1 = 0\).
- For each equation, identify all solutions within the interval. For ours, that's angles in \([0, 2\pi)\).
- Utilize knowledge of the sine function and the unit circle to pinpoint exact angle solutions.
