solve algebraically, express the roots in exact form

2sinxcosx-1=0
I got:
2sinxcosx=1
sin^2x=1
what do I do next?

thanks in advance

To continue solving the equation 2sin(x)cos(x) - 1 = 0, you have correctly simplified it to 2sin(x)cos(x) = 1 and sin^2(x) = 1. Now, let's continue step-by-step to find the exact roots algebraically:

1. Since sin^2(x) = 1, there are two possibilities:
a) sin(x) = 1
b) sin(x) = -1

2. For sin(x) = 1, you can solve for x by taking the inverse sine (sin^(-1)) of both sides:
x = sin^(-1)(1)

3. For sin(x) = -1, you can solve for x by taking the inverse sine (sin^(-1)) of both sides:
x = sin^(-1)(-1)

4. However, we need to consider the range of the inverse sine function, which is typically -π/2 ≤ sin^(-1)(x) ≤ π/2. Since sin(x) = 1 has multiple solutions, we need to find all solutions within the range.

5. For sin^(-1)(1), we know that sin(π/2) = 1. Since sin(x) is a periodic function with a period of 2π, other solutions can be found by adding multiples of 2π to π/2:
x = π/2 + 2πn, where n is an integer.

6. Similarly, for sin^(-1)(-1), we know that sin(-π/2) = -1. Other solutions can be found by adding multiples of 2π to -π/2:
x = -π/2 + 2πn, where n is an integer.

Therefore, the exact roots of the equation 2sin(x)cos(x) - 1 = 0, expressed in exact form, are:
x = π/2 + 2πn, for n being any integer (for sin(x) = 1)
x = -π/2 + 2πn, for n being any integer (for sin(x) = -1)