1. a) (3x² -9) /(2x² -4x - 6) > 0 ⇔ 3(x² -3)/[2(x² - 2x -3)] > 0
3(x² - 3)/[2(x-3)(x+1)] >0
I. x² - 3 > 0 x > √3 x - 3> 0 x > 3 x + 1 > 0 x > - 1 ⇒ x> 3
x ∈ (3,+∞)
II. x²-3 < 0 x < √3 x - 3 <0 x < 3 x+1>0 x > - 1
x ∈ (-1,3)
III. x² - 3 <0 x < √3 x-3>0 x > 3 x+1<0 x< -1
x ∈ (-∞, -1) ∪ (3,+∞)
⇒ x ∈ (-∞, -1) ∪ (-1,3) ∪ (3,+∞)
b) 2x-1 ≥ x² - 2x + 3 ⇔ x² - 4x + 4 ≤ 0 (x -2)² = 0 x = 2
2. sin x - cosx = 1/2 sin²x - 2sinx cosx +cos²x = 1/4 2sinxcosx = 1-1/4
a) sin 2x = 2sinxcosx = 3/4
b)cos2x = cos²x - sin²x = (cosx - sinx)(cosx + sinx) = -1/2(cosx + sinx) =
=-1/2 ·√5/2 = - √5
( sinx + cosx)² = 1 + 3/4 = 5/4 ⇒ sinx + cosx = √5 /2
c) tg2x = sin2x/cos2x = 3/4 ·2/√5 = 3/(2√5) = 3√5/10
