sin(4x) = 1/2
Understand the Problem
The question is asking to solve the equation sin(4x) = 1/2 for the variable x. This involves using the properties of the sine function and potentially applying inverse sine functions and considering the periodic nature of trigonometric functions.
Answer
The solutions are $x = \frac{\pi}{24} + \frac{k\pi}{2}$ and $x = \frac{5\pi}{24} + \frac{k\pi}{2}$.
Answer for screen readers
The solutions to the equation $\sin(4x) = \frac{1}{2}$ are: $$ x = \frac{\pi}{24} + \frac{k\pi}{2} \quad \text{and} \quad x = \frac{5\pi}{24} + \frac{k\pi}{2} $$
Steps to Solve
- Identify the sine function values
The equation we need to solve is given by: $$ \sin(4x) = \frac{1}{2} $$
We know that the sine function equals $\frac{1}{2}$ at specific standard angles.
- Determine the angles
The sine function equals $\frac{1}{2}$ at: $$ 4x = \frac{\pi}{6} + 2k\pi $$ and $$ 4x = \frac{5\pi}{6} + 2k\pi $$
where $k$ is any integer representing the periodic nature of the sine function.
- Solve for x in each case
For the first equation: $$ 4x = \frac{\pi}{6} + 2k\pi $$ Dividing by 4 gives: $$ x = \frac{\pi}{24} + \frac{k\pi}{2} $$
For the second equation: $$ 4x = \frac{5\pi}{6} + 2k\pi $$ Again, dividing by 4 gives: $$ x = \frac{5\pi}{24} + \frac{k\pi}{2} $$
- Combine the solutions
The general solutions for $x$ are: $$ x = \frac{\pi}{24} + \frac{k\pi}{2} $$ and $$ x = \frac{5\pi}{24} + \frac{k\pi}{2} $$
where $k$ is an integer.
The solutions to the equation $\sin(4x) = \frac{1}{2}$ are: $$ x = \frac{\pi}{24} + \frac{k\pi}{2} \quad \text{and} \quad x = \frac{5\pi}{24} + \frac{k\pi}{2} $$
More Information
The solutions represent all angles where the sine function equals $\frac{1}{2}$, repeating every $\pi/2$ radians due to the periodic nature of the sine wave. This means that there are infinitely many values of $x$ that satisfy this equation, depending on the integer $k$.
Tips
- Forgetting to include the periodic nature of the sine function, which leads to missing additional solutions.
- Not considering negative integers for $k$, which can also yield valid solutions.
- Misidentifying the angles where $\sin(x) = \frac{1}{2}$.
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