If sin x + cos x = -1/5, and 3/4*pi <= x <= pi, then find cos 2x.
sin x + cos x = -1/5 square both sides
sin^2 x + 2sinxcosx + cos^2x = 1/25
1 + 2sin xcosx = 1/25
2sinxcosx = -24/25
sin (2x) = -24/25
cos (2x) = ±√ [ 1 - sin^2(2x) ]
cos (2x) = ±√[ 1 - (-24/25)^2 ]
cos (2x) = ±√[ 625 - 576]/ 25
cos (2x) = ± (7/25)
Since x = [ (3/4)pi , pi ] then
arccos [ 7/25] = x ≈ 1.287 rads
Then dividing by 2, x ≈ .6435 rads......reject
arccos [ -(7/25) ] = x ≈ 1.85 rads
Then dividing by 2, x ≈ .927 rads .....reject
And
arccos [ 7/25] = 2pi - 1.287 = 4.996 rads
Then dividing by 2, x ≈ 2.498 rads.....accept
So cos 2x = 7/25