Решение смотри в приложении
2-3(2x+2)=5-4x
2-6x-6=5-4x
-6x+4x=5-2+6
-2x=9
x=-4.5
Решение
(3cos2x+5cosx-1)/√(- ctgx) = 0
<span>{3cos2x+5cosx-1 = 0
</span>{√(- ctgx) ≠ 0, ctgx ≠ 0, x ≠ π/2+ πk, k ∈ Z
<span>
3cos2x+5cosx-1 = 0</span>
3*(2cos²x - 1) + 5cosx - 1 = 0
6cos²x + 5cosx - 4 = 0
cosx = t, I t I ≤ 1
6t² + 5t - 4 = 0
D = 25 + 4*6*4 = 121
t₁ = (- 5 - 11)/12
t₁ = - 16/12 = - 4/3, не удовлетворяет условию I <span>t I ≤ 1
</span>t₂ = <span>(- 5 + 11)/12
</span>t₂ = 1/2
cosx = 1/2
x = (+-) arccos(1/2) + 2πn, n ∈ Z
x = (+-) *(π/3) + 2πn, n ∈ <span>Z
</span>
Ответ: x = (+-) *(π/3) + 2πn, n ∈ <span>Z, </span>