To solve the equation sin(2x) * sin(x) = 0, we need to find the values of x that make either sin(2x) or sin(x) equal to zero.
### Step 1: Set Each Factor to Zero
1. sin(2x) = 0
- The sine function equals zero at integer multiples of π:
- So, we set 2x = nπ, where n is an integer.
- This gives us x = (nπ)/2.
Now we find the values of x in the interval [0, 2π]:
- For n=0: x = 0
- For n=1: x = π/2
- For n=2: x = π
- For n=3: x = 3π/2
- For n=4: x = 2π
Thus, the solutions from sin(2x) = 0 are: x = 0, π/2, π, 3π/2, 2π.
2. sin(x) = 0
- Again, sine equals zero at integer multiples of π:
- So, we set x = mπ, where m is an integer.
Now we find the values of x in the interval [0, 2π]:
- For m=0: x = 0
- For m=1: x = π
- For m=2: x = 2π
The solutions from sin(x) = 0 are: x = 0, π, 2π.
### Step 2: Combine Solutions
Now we combine all the unique solutions found:
- x = 0
- x = π/2
- x = π
- x = 3π/2
- x = 2π
### Summary of Solutions
To summarize, the solutions of the equation sin(2x) * sin(x) = 0 in the interval from 0 to 2π are:
- x = 0
- x = π/2
- x = π
- x = 3π/2
- x = 2π