How do you solve #2sin^2(2x)=1#?

2 Answers
Noah G
Sep 12, 2016

#sin^2(2x) = 1/2#

#sin(2x) = +-1/sqrt(2)#

#2x = +-sin^-1(1/sqrt(2))#

#2x = 135˚, 45˚, 225˚, 315˚#

#x = 67.5˚, 22.5˚, 112.5˚ and 157.5˚#

Hopefully this helps!

Nghi N.
Sep 13, 2016

#pi/8. (3pi)/8, (5pi)/8, (7pi)/8#

Explanation:

#2sin^2 2x = 1#
#sin^2 2x = 1/2#
#sin 2x = +- 1/sqrt2 = +- sqrt2/2#
Trig table and unit circle -->

a. #sin 2x = sqrt2/2# --> #2x = pi/4, and 2x = (3pi)/4#
When #2x = pi/4# --> #x = pi/8#
When #2x = (3pi)/4# --> #x = (3pi)/8#

b. #sin 2x = - sqrt2/2# --> #2x = - (3pi)/4#, and #x = - pi/4#
Arc #(-3pi)/4# is co-terminal to arc #(5pi)/4#
Arc #(-pi/4)# is co-terminal to arc #(7pi)/4#
When #2x = (5pi)/4# --> #x = (5pi)/8#
When #2x = (7pi)/4# --> #x = (7pi)/8#
Answers for #(0, 2pi)#
#pi/8, (3pi)/8, (5pi)/8, (7pi)/8#

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