To find all solutions to the equation 2sin(2x)sin(4x) = 0 in the interval [0, 2π], we first recognize that the product is zero when either factor is zero.
This gives us two separate equations to solve:
- 1. 2sin(2x) = 0
- 2. sin(4x) = 0
Solving 2sin(2x) = 0:
We divide both sides by 2 (noting that this does not affect the equality) to get:
sin(2x) = 0
The sine function is zero at integer multiples of π:
2x = nπ for n ∈ ℤ.
Therefore, we solve for x:
x = nπ/2
Now, to find the values of x in the interval [0, 2π], we can substitute values for n:
- n = 0: x = 0
- n = 1: x = π/2
- n = 2: x = π
- n = 3: x = 3π/2
- n = 4: x = 2π
The solutions from this equation are: x = 0, π/2, π, 3π/2, 2π.
Solving sin(4x) = 0:
Similar to the previous case:
4x = mπ for m ∈ ℤ.
Now, solving for x gives:
x = mπ/4
We again substitute integer values for m to find solutions in the interval [0, 2π]:
- m = 0: x = 0
- m = 1: x = π/4
- m = 2: x = π/2
- m = 3: x = 3π/4
- m = 4: x = π
- m = 5: x = 5π/4
- m = 6: x = 3π/2
- m = 7: x = 7π/4
- m = 8: x = 2π
The solutions from this equation are: x = 0, π/4, π/2, 3π/4, π, 5π/4, 3π/2, 7π/4, 2π.
Combining Both Solutions:
Now, let’s combine all the solutions we found:
- x = 0
- x = π/4
- x = π/2
- x = 3π/4
- x = π
- x = 5π/4
- x = 3π/2
- x = 7π/4
- x = 2π
Therefore, the complete solution set for the equation 2sin(2x)sin(4x) = 0 in the interval [0, 2π] is:
- x = 0, π/4, π/2, 3π/4, π, 5π/4, 3π/2, 7π/4, 2π