Solve this over the interval [0,2pi):
sin(3x)=(square root of 2)/2
For sin(α)=(√2)/2,
Thus α=π/4 or 3π/4.
To find the complete solution of x between 0 and 2π, we set:
3x = π/4 + 2kπ, where k is an integer, or
3x = 3π/4 + 2kπ
For the first case,
x=π/12, π/12+(2π/3) or π/12+(4π/3)
=π/12, 3π/4, 17π/12
For the second case,
x=π/4, π/4+(2π/3) or π/4+(4π/3)
=π/4, 11π/12, 19π/12
So the solution set is:
x={π/12, π/4, 3π/4, 11π/12, 17π/12, 19π/12 }
Graph the function sin(3x) and convince yourself that there are indeed six roots to the given equation.
To solve the equation sin(3x) = √2/2 over the interval [0, 2π), you can follow these steps:
1. Start by finding the inverse of the sine function. Since sin(x) = √2/2, you can find the angle x in the unit circle where the sine is √2/2. This angle is π/4 (or 45°).
2. Now, you need to find all the solutions in the given interval [0, 2π) that satisfy sin(3x) = √2/2.
3. Divide the interval [0, 2π) into two parts: [0, π) and [π, 2π).
4. In the first interval [0, π), find the angles that satisfy sin(3x) = √2/2. Since sin(x) is a periodic function, you can find the solutions by solving 3x = π/4. Divide both sides by 3 to get the value of x: x = π/12.
5. In the second interval [π, 2π), find the angles that satisfy sin(3x) = √2/2. Again, you solve 3x = π/4. However, since you are in the second interval, you need to add π to the solutions obtained in the first interval: x = π/12 + π = (13π)/12.
6. Therefore, the solutions in the interval [0, 2π) that satisfy sin(3x) = √2/2 are x = π/12 and x = (13π)/12.
So, the solutions in the interval [0, 2π) for the equation sin(3x) = √2/2 are x = π/12 and x = (13π)/12.