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

The Correct Option is D

To determine the number of solutions for the equation \(4 \sin^2 x - 4 \cos^3 x + 9 - 4 \cos x = 0, \, x \in [-2\pi, 2\pi]\), we will analyze and solve the equation step-by-step.

Firstly, let's rewrite the equation:

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

Using the Pythagorean identity \(\sin^2 x = 1 - \cos^2 x\), we substitute it into the equation:

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

Simplifying further:

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

This results in:

\(13 - 4 \cos^2 x - 4 \cos^3 x - 4 \cos x = 0\)

Reorganizing terms gives:

\(-4 \cos^3 x - 4 \cos^2 x - 4 \cos x + 13 = 0\)

Now let \(y = \cos x\). This transforms the equation into:

\(-4y^3 - 4y^2 - 4y + 13 = 0\)

This is a cubic polynomial in \(y\), for which we need to find real roots within the range \(y \in [-1, 1]\) because \(\cos x\) can only take values within this range.

Observe the function:

\(f(y) = -4y^3 - 4y^2 - 4y + 13\)

Checking at the boundaries of the interval \([-1, 1]\):

• For \(y = -1\), \(f(-1) = -4(-1)^3 - 4(-1)^2 - 4(-1) + 13 = 4 - 4 + 4 + 13 = 17\)
• For \(y = 1\), \(f(1) = -4(1)^3 - 4(1)^2 - 4(1) + 13 = -4 - 4 - 4 + 13 = 1\)

Since \(f(-1) = 17\) and \(f(1) = 1\), and both are positive, it indicates there are no sign changes within \(y = -1\) to \(y = 1\). Therefore, there are no real roots within this interval.

Hence, there are no solutions for the equation \(4 \sin^2 x - 4 \cos^3 x + 9 - 4 \cos x = 0\) in the interval \(x \in [-2\pi, 2\pi]\).

Therefore, the correct answer is 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.