Question

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

Answer 1

`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!

Answer 2

`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`