Question

Solve this over the interval [0,2pi):

sin(3x)=(square root of 2)/2

Answer 1

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.