Asked by Renee
Solve this over the interval [0,2pi):
sin(3x)=(square root of 2)/2
sin(3x)=(square root of 2)/2
Answers
Answered by
MathMate
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.
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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.