Find all solutions in the interval [0, 2π).
sin2x - cos2x = 0
sin(2x) - cos(2x) = 0 => sin(2x) = cos(2x) => tan(2x) = 1
2x = π/4, 5π/4, 9π/4, 13π/4
x = π/8, 5π/8, 9π/8, 13π/8
sin2x=cos2x
u=2x
sinu=cosu
u must be pi/4 and 5pi/4 and 9pi/4 and 13pi/4
since those are u, and we are finding x, solve for x for each of these u's
pi/8, 5pi/8, 9pi/8, 13pi/8
these are all the value in the interval.
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sin(2x) - cos(2x) = 0 => sin(2x) = cos(2x) => tan(2x) = 1
2x = π/4, 5π/4, 9π/4, 13π/4
x = π/8, 5π/8, 9π/8, 13π/8
sin2x=cos2x
u=2x
sinu=cosu
u must be pi/4 and 5pi/4 and 9pi/4 and 13pi/4
since those are u, and we are finding x, solve for x for each of these u's
pi/8, 5pi/8, 9pi/8, 13pi/8
these are all the value in the interval.