Решение
12sin^2x-8cosx+1=0
12(1 - cos²x) - 8cosx + 1 = 0
12 - 12cos²x - 8cosx + 1 = 0
- 12cos²x - 8cosx + 13 = 0
12cos²x + 8cosx - 13 = 0
I cosx I ≤ 1
cosx = t
12t² + 8t - 13 = 0
D = 64 + 4*12*13 = 688
t₁ = (-8 - 4√43)/24
t₁ = ( - 2 - √43) / 6 не удовлетворяет условию I cosx I ≤ 1
t₂ = (- 2 + √43) / 6
cosx = (- 2 + √43) / 6
x = (+ -)arccos((- 2 + √43) / 6) + 2πk, k∈Z