1) cos3x + cosx = 0
2cos(3x + x)/2*cos(3x - x)/2 = 0
cos2x * cosx = 0
a) cos2x = 0
2x = π/2 + πn, n∈Z
x₁ = π/4 + πn/2, n ∈Z
n = - 1
x = π/4 - π/2 = - π/4 ∈ [- π/2;π/2]
n = - 2
x = π/4 - π = - 3π/4 ∉ [- π/2;π/2]
n = 0
x = π/4 <span>∈ [- π/2;π/2]
n = 1
x = </span>π/4 + π/2 = 3π/4 ∉ [- π/2;π/2]
n = 2
π/4 + π = 5π/4 ∉ <span>[- π/2;π/2]
</span>Ответ: - π/4; π/4
b) cosx = 0
x = π/2 + πk, k∈Z
k = - 1
x = π/2 - π = - π/2 ∈ <span>[- π/2;π/2]
</span>k = 0
x = π/2 ∈ <span>[- π/2;π/2]
</span>k = 1
x = π/2 + π = 3π/2 ∉ <span>[- π/2;π/2]
</span>Ответ: - π/2; π/2
2) 2sin² x - sin2x = cos2x
2sin²x
- 2sinxcosx - (2cos²x - 1) = 0
<span>2sin²x
- 2sinxcosx - 2cos²x + sin²x + cos²x = 0
</span>3sin²x
- 2sinxcosx – cos²<span>x = 0 / делим на cos²x ≠ 0
</span><span>3tg²x
- 2tgx - 1 = 0
</span><span>D = 4
+ 4*3*1 = 16
</span><span>1) tgx = (2 - 4)/6
</span><span>tgx = - 1/3
</span>x<span>₁ = - arctg(1/3) + πn, n∈Z
</span><span>tgx = ( 2 + 4)/6
</span><span>tgx = 1
</span>x₂ = π/4 + πk, k∈Z