Question

Find the number of solutions of the equation x3 + 2x2 + 5x + 2cos x = 0 in [0, 2π]

Answer 1

We have x3 + 2x2 + 5x + 2cosx = 0

Let f(x) = x3 + 2x2 + 5x + 2cosx

⇒ f'(x) = 3x2 + 4x + 5 – 2sinx

Let g(x) = 3x2 + 4x + 5 and h(x) = 2sinx

Max value of g(x) is – (16 – 60)/6 = 44/6 = 22/3

and max value of h (x) is 2.

Thus, f'(x) > 0

⇒ f(x) is strictly increasing function

Also, f(0) = 2 > 0 and f(2π) > 0

Therefore f(x) has no real root.