Trigonometry
posted by Grace .
How do I find all solutions in the interval [0,2Pi):
sin2x=cos2x
answer choices are:
a.)Pi/8
b.) Pi/8, 5Pi/8, 9Pi/8, 13Pi/8

The solution is tan 2x = 1
2x = arctan 1
2x = pi/4 or (5/4) pi or (9/4) pi or (13 pi/8). You can ignore higher values because of the 2 pi limit on x.
Solve for x
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