Given: \( 3 \tan 4 x=-2 \cos 4 x \) 5.5.1 Without using a calculator, show that \( \sin 4 x=-0,5 \) is the only solution to the above equation. 5.5.2 Hence, determine the general solution of \( x \) in the equation \[ 3 \tan 4 x=-2 \cos 4 x \] [28]

To start with \( 3 \tan 4x = -2 \cos 4x \), we can rewrite the tangent function using the sine and cosine definitions: \[ 3 \frac{\sin 4x}{\cos 4x} = -2 \cos 4x. \] Multiplying both sides by \( \cos 4x \) (assuming \( \cos 4x \neq 0 \)) leads to: \[ 3 \sin 4x = -2 \cos^2 4x. \] Using the Pythagorean identity \( \cos^2 4x = 1 - \sin^2 4x \) lets us express everything in terms of sine: \[ 3 \sin 4x = -2 (1 - \sin^2 4x), \] which simplifies to: \[ 3 \sin 4x = -2 + 2 \sin^2 4x. \] Rearranging this gives us a quadratic equation: \[ 2 \sin^2 4x - 3 \sin 4x - 2 = 0. \] We can factor or use the quadratic formula \( \sin 4x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 2, b = -3, c = -2 \): Calculating the discriminant: \[ (-3)^2 - 4 \cdot 2 \cdot (-2) = 9 + 16 = 25. \] So, we find: \[ \sin 4x = \frac{3 \pm 5}{4}. \] This results in: 1. \( \sin 4x = \frac{8}{4} = 2 \) (not a valid sine value) 2. \( \sin 4x = \frac{-2}{4} = -0.5 \). Thus, \( \sin 4x = -0.5 \) is the only valid solution. For the second part, we need to find the general solution for \( 4x \): \[ 4x = \arcsin(-0.5) + 2k\pi, \] where \( k \) is any integer. The solutions can be derived from the unit circle, where \( \sin \theta = -0.5 \) at \( \theta = \frac{7\pi}{6} + 2k\pi \) and \( \theta = \frac{11\pi}{6} + 2k\pi \). Thus, we have: \[ 4x = \frac{7\pi}{6} + 2k\pi \quad \text{or} \quad 4x = \frac{11\pi}{6} + 2k\pi. \] Dividing each equation by 4 gives: \[ x = \frac{7\pi}{24} + \frac{k\pi}{2} \quad \text{or} \quad x = \frac{11\pi}{24} + \frac{k\pi}{2}. \] Hence, the general solutions for \( x \) are: \[ x = \frac{7\pi}{24} + \frac{k\pi}{2} \quad \text{and} \quad x = \frac{11\pi}{24} + \frac{k\pi}{2}, \quad k \in \mathbb{Z}. \]