To solve the equation 0 = 2sin²(x)cos²(x), we can start by simplifying the equation.
Notice that 2sin²(x)cos²(x) can be rewritten using the double angle identity for sine. The identity states:
- sin(2x) = 2sin(x)cos(x)
Therefore, we can express 2sin²(x)cos²(x) as:
- 2sin²(x)cos²(x) = (sin(2x))² / 2
Now, the equation becomes:
- 0 = (sin(2x))² / 2
Multiplying both sides by 2 yields:
- 0 = (sin(2x))²
For the above equation to hold true, we must have:
- sin(2x) = 0
The sine function is equal to zero at various points, specifically:
- 2x = nπ, where n is an integer.
This gives us:
- x = nπ / 2
We are interested in solutions for x over the interval [0, π]. Let’s find the values of n that fit within this interval:
- For n = 0: x = 0
- For n = 1: x = π / 2
- For n = 2: x = π
- For n = 3: x = 3π / 2 (not in the interval)
Thus, the solutions to the equation 0 = 2sin²(x)cos²(x) on the interval [0, π] are:
- x = 0
- x = π / 2
- x = π
In conclusion, the final solutions are:
- x = 0
- x = π / 2
- x = π