Question

Question #61687

Answer 1

`sin^2x-cos^2x=sqrt2/2`

`=>-cos2x=sqrt2/2=1/sqrt2`

`=>cos2x=-1/sqrt2=-cos(pi/4)`

`cos2x=cos(pi-pi/4)=cos((3pi)/4)`

`=>2x=2npi±((3pi)/4)`

`=>x=npi±((3pi)/8)," where "n in ZZ`

Answer 2

(3pi)/8 + kpi
(5pi)/8 + kpi

Explanation:

Use trig identity: `(sin^2 x - cos^2 x = - cos 2x)`
`-cos 2x = sqrt2/2`
`cos 2x = - sqrt2/2`
Unit circle and trig table of special arcs -->
`cos 2x = -sqrt2/2` gives 2 solutions arcs for 2x:
`2x = (3pi)/4` and `2x = (5pi)/4` (co-terminal to `(-3pi)/4`)

a. `2x = (3pi)/4 + 2kpi`
`x = (3pi)/8 + kpi`

b. `2x = (5pi)/4 + 2kpi`
`x = (5pi)/8 + kpi`