Chapter 6: Problem 7
a) Solve algebraically \(\sin 2 x=0.5\) \(\mathbf{0} \leq x<2 \pi\) b) Solve the equation from part a) using a different method.
Short Answer
Expert verified
x = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}
Step by step solution
01
Understand the Equation
The given equation is \(\sin 2x = 0.5\). This is a trigonometric equation.
02
Use the Inverse Sine Function
To solve \(\sin 2x = 0.5\), take the inverse sine (arcsine) of both sides to get \[2x = \sin^{-1}(0.5)\].
03
Determine Principal Values
The value of \(\sin^{-1}(0.5)\) is \(\frac{\pi}{6}\). Therefore, \([2x = \frac{\pi}{6}, \pi - \frac{\pi}{6})\).
04
Solve for x
Divide each of the possible values for 2x by 2: \[x = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}\].
05
Check within the Interval
Verify that all solutions are within the interval 0 \(\leq x < 2\pi\). All solutions \(\frac{\pi}{12}, \frac{5\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}\) fall within the range.
06
Solve using Graphical Method
Another method to solve \(\sin 2x = 0.5\) is by sketching the graphs of \(\sin 2x\) and \(\frac{1}{2}\) and finding their points of intersection.
07
Find Intersections
Plot the graphs of \(\sin 2x\) and \(\frac{1}{2}\). The points of intersection (in radians) will be the same as those from the algebraic method: \(\frac{\pi}{12}, \frac{5\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Inverse Sine Function
The inverse sine function, often denoted as \(\text{sin}^{-1}(x)\), is used to find the angle whose sine is a given number. In the given problem, we need to solve \(\text{sin}(2x) = 0.5\). To proceed algebraically, we use the inverse sine to isolate the angle: \(\text{sin}^{-1}(0.5)\).
Remember, the sine function returns values between -1 and 1, so make sure that the right side of your equation falls within this interval. Otherwise, the equation would have no solutions.
Remember, the sine function returns values between -1 and 1, so make sure that the right side of your equation falls within this interval. Otherwise, the equation would have no solutions.
Trigonometric Graphing
Graphing trigonometric functions is an excellent way to visualize their behavior and solve equations. In the graphical method, we plot the functions \(\text{sin}(2x)\) and \(\frac{1}{2}\) on the same set of axes.
Look where the two graphs intersect—these intersections represent the solutions to the equation. It is especially useful for understanding periodic behavior and can help verify algebraic solutions.
Look where the two graphs intersect—these intersections represent the solutions to the equation. It is especially useful for understanding periodic behavior and can help verify algebraic solutions.
Principal Values
Principal values refer to the main solutions of an inverse trigonometric function within its restricted domain. For \(\text{sin}^{-1}(x)\), the principal values lie between \(-\frac{\text{π}}{2}\) and \(\frac{\text{π}}{2}\).
When solving \(\text{sin}(2x) = 0.5\), \(\text{sin}^{-1}(0.5)\) yields \(\frac{\text{π}}{6}\). However, sine being periodic means there are also other values for 2x such that \(\text{sin}(2x) = 0.5\). These can be found using \(\text{π} - \text{θ}\).
When solving \(\text{sin}(2x) = 0.5\), \(\text{sin}^{-1}(0.5)\) yields \(\frac{\text{π}}{6}\). However, sine being periodic means there are also other values for 2x such that \(\text{sin}(2x) = 0.5\). These can be found using \(\text{π} - \text{θ}\).
Interval Checking
When solving trigonometric equations, it’s crucial to ensure that your solutions lie within the specified interval, here \(0 \leq x < 2 \pi\).
Once you've obtained possible solutions, check if each one fits the interval. If it does, it's a valid solution. Otherwise, discard it. In the problem, after dividing by 2, we check values \(\frac{\text{π}}{12}, \frac{5\text{π}}{12}, \frac{13\text{π}}{12}, \frac{17\text{π}}{12}\)—all of which are valid within the given interval.
Once you've obtained possible solutions, check if each one fits the interval. If it does, it's a valid solution. Otherwise, discard it. In the problem, after dividing by 2, we check values \(\frac{\text{π}}{12}, \frac{5\text{π}}{12}, \frac{13\text{π}}{12}, \frac{17\text{π}}{12}\)—all of which are valid within the given interval.