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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