Final answer:
The equation cos²x + 3cosx + 1 = 0 has one solution for 0 < x < 2π.
Step-by-step explanation:
To determine the number of solutions to the equation cos²x + 3cosx + 1 = 0 for 0 < x < 2π, we can use the quadratic formula. For an equation of the form ax²+ bx + c = 0, the quadratic formula is x = (-b ± sqrt(b² - 4ac))/(2a). In this equation, a = 1, b = 3, and c = 1.
Plugging these values into the quadratic formula, we get x = (-3 ± sqrt(9 - 4(1)(1)))/(2(1)). Simplifying further, we have x = (-3 ± sqrt(5))/2. Since -3 ± sqrt(5) is approximately -5.24 and 0.24 is greater than 0, we can discard the negative root. Therefore, x ≈ (-3 + sqrt(5))/2.
So, there is one solution to the equation cos²x + 3cosx + 1 = 0 for 0 < x < 2π.