How do you solve #sqrt(2)sin2x=1# ?

1 Answer
Jake M.
Apr 8, 2018

#x in {pi/8 + pi n, (3pi)/8 + pi n} forall n in mathbb{Z} #

Explanation:

#sqrt2 sin2x = 1 implies sin2x = 1/sqrt2 = sqrt2/2#

So #2x = sin^(-1)(sqrt2/2) #

From our knowledge of the unit circle, we know that
#sin^(-1)(sqrt2/2) = pi/4, (3pi)/4# and all times around the circle, i.e.

#2x in { pi/4 + 2pi n, (3pi)/4 + 2pi n } forall n in mathbb{Z} #

Therefore,
#x in {pi/8 + pi n, (3pi)/8 + pi n} forall n in mathbb{Z} #

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