Solve for x on the interval (0, 2pi]
(sin2x + cos2x)^2 = 1
sin^2 2x + 2sin(2x) cos(2x) + cos^2 2x = 1
1 + sin (4x) = 1
sin 4x = 0
4x = 0, π, 2π, 3π ....
x = 0 , π/4, π/2, 3π/4, π , 5π/4, 3π/2, 7π/4, 2π
choose the answers according to your notation (0,2π]
To solve the equation (sin2x + cos2x)^2 = 1, we can follow these steps:
1. Expand the square of the binomial:
sin^2(2x) + 2sin(2x)cos(2x) + cos^2(2x) = 1
2. Use the trigonometric identity: sin^2(θ) + cos^2(θ) = 1
Rearranging and substituting:
sin^2(2x) + cos^2(2x) = 1 - 2sin(2x)cos(2x)
3. Substitute the identity into the expanded equation:
1 - 2sin(2x)cos(2x) + 2sin(2x)cos(2x) = 1
4. Simplify the equation:
-2sin(2x)cos(2x) + 2sin(2x)cos(2x) = 0
5. Notice that the term -2sin(2x)cos(2x) cancels out, leaving us with:
0 = 0
6. This equation is always true, regardless of the value of x. Therefore, any value of x within the given interval (0, 2pi] is a solution.
Thus, the solution to the equation (sin2x + cos2x)^2 = 1 on the interval (0, 2pi] is x ∈ (0, 2pi].