Question

The number of solutions of the equation $4 \sin^2 x - 4 \cos^3 x + 9 - 4 \cos x = 0, \, x \in [-2\pi, 2\pi]$is:

• A. 1
• B. 3
• C. 2
• D. 0

Answer 1

Answer: (D)

0

Answer 2

Given equation:

$4\sin^2 x - 4\cos^3 x + 9 - 4\cos x = 0$

We use the identity:

$\sin^2 x = 1 - \cos^2 x$

Substituting this in the equation:

$4(1 - \cos^2 x) - 4\cos^3 x + 9 - 4\cos x = 0$

Simplifying:

$4 - 4\cos^2 x - 4\cos^3 x + 9 - 4\cos x = 0$

Combining like terms:

$13 - 4\cos^3 x - 4\cos^2 x - 4\cos x = 0$

Factoring out $-4$:

$-4(\cos^3 x + \cos^2 x + \cos x - \frac{13}{4}) = 0$

Therefore, we need to solve:

$\cos^3 x + \cos^2 x + \cos x - \frac{13}{4} = 0$

After analyzing the roots of this equation within the interval $x \in [-2\pi, 2\pi]$, we observe there are no real solutions that satisfy the equation.

Conclusion: The number of solutions is 0.