Question

сosx + cos2x + cos3x = 0

Answer 1

1. cosx + cos2x + cos3x = 0cos2x + (cosx + cos3x) = 0cos2x + 2 cos (3x + x / 2) cos (3x - x / 2) = 0cos2x + 2 cos2xcosx = 0cos2x (1 + 2cosx) = 0cos2x = 0 или 1 + 2cosx = 0 1) cos2x = 0cos2x = cosπ/2 + πn2x = π/2 + πnx = π/4 + πn/2; n принадлежит Z. 2) 1 + 2 cosx = 0cos x = -1/2x1 = 2π/3 + 2πn;x2 = -2π/3 + 2πn;n принадлежит Z.

Answer 2

(cosx+cos3x)+cos2x=0
2cos2x*cosx+cos2x=0
cos2x(2cosx+1)=0
cos2x=0
2x=±arccos 0 +2πn, n ∈ Z
2x=±π/2+πn, n ∈ Z
x=π/4+πn/2, n ∈ Z
2+cosx =0
2cosx+1=0
cosx=-1/2
x=±2π/3 + 2πn, n ∈ Z

Ответ: π/4+πn/2, ±2π/3 + 2πn