To solve the equation sin(2x) sin(x) = 0, we start by recognizing that a product of two factors is zero if at least one of the factors is zero. Therefore, we need to set each factor equal to zero separately:
1. Setting sin(2x) = 0:
The sine function equals zero at integer multiples of π:
sin(2x) = 0 → 2x = nπ, where n is an integer.
Dividing both sides by 2 gives us:
x = nπ/2.
For 0 ≤ x < 2π, the integer values for n that will keep x in this range are n = 0, 1, 2, 3, and 4. Therefore:
- n = 0: x = 0
- n = 1: x = π/2
- n = 2: x = π
- n = 3: x = 3π/2
- n = 4: x = 2π
Thus, the solutions from this equation are:
- x = 0
- x = π/2
- x = π
- x = 3π/2
- x = 2π
2. Setting sin(x) = 0:
Similarly, the sine function equals zero at integer multiples of π:
sin(x) = 0 → x = mπ, where m is an integer.
For 0 ≤ x < 2π, the integer values for m that will keep x in this range are m = 0, 1, and 2. Therefore:
- m = 0: x = 0
- m = 1: x = π
- m = 2: x = 2π
Thus, the solutions from this equation are:
- x = 0
- x = π
- x = 2π
Final Solutions:
Combining both sets of solutions, we have:
- x = 0
- x = π/2
- x = π
- x = 3π/2
- x = 2π
So, the complete set of solutions for the equation sin(2x) sin(x) = 0 in the interval [0, 2π) is:
- x = 0
- x = π/2
- x = π
- x = 3π/2