Пишу).........................
<h3>3cos²x + 2sinx•cosx = 0</h3><h3>cosx•( 3cosx + 2sinx ) = 0</h3><h3>1) cosx = 0 ⇔ x = π/2 + πn, n ∈ Z</h3><h3>2) 3cosx + 2sinx = 0 ⇒ разделим обе части на cosx ≠ 0</h3><h3>2tgx + 3 = 0 ⇔ tgx = - 3/2 ⇔ x = - arctg(3/2) + πn, n ∈ Z</h3><h3><u><em>ОТВЕТ: π/2 + πn ; - arctg(3/2) + πn, n ∈ Z</em></u></h3><h3><u><em /></u></h3>
(150-х) :2=65
150-х=65x2
150-x=130
x=150-130
x=20
48-16:2=40
65-(20+15)=30
50+3*4=62
28-24:3=20