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 show that \( \sin 4x = -0.5 \) is the only solution to the equation \( 3 \tan 4x = -2 \cos 4x \), we start with the identity \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). So, we can rewrite the left side: \[ 3 \frac{\sin 4x}{\cos 4x} = -2 \cos 4x \] Multiplying both sides by \( \cos 4x \) (assuming \( \cos 4x \neq 0 \)), we have: \[ 3 \sin 4x = -2 \cos^2 4x \] Using the identity \( \cos^2 4x = 1 - \sin^2 4x \), we can substitute: \[ 3 \sin 4x = -2 (1 - \sin^2 4x) \] Rearranging this gives: \[ 3 \sin 4x + 2 - 2 \sin^2 4x = 0 \] Let's denote \( y = \sin 4x \). The equation becomes: \[ 2y^2 - 3y - 2 = 0 \] We can solve this quadratic using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2, b = -3, c = -2 \): \[ y = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm 5}{4} \] This gives us: \[ y_1 = \frac{8}{4} = 2 \quad \text{and} \quad y_2 = \frac{-2}{4} = -0.5 \] Since \( \sin 4x \) can only take values in the range of \([-1, 1]\), the solution \( y_1 = 2 \) is extraneous. Therefore, the only valid solution is \( y = \sin 4x = -0.5 \). Next, to find the general solution for \( x \) given \( \sin 4x = -0.5 \): The solutions for \( \sin \theta = -0.5 \) occur when: \[ 4x = \frac{7\pi}{6} + 2k\pi \quad \text{or} \quad 4x = \frac{11\pi}{6} + 2k\pi \quad (k \in \mathbb{Z}) \] Dividing each of these equations by 4 gives us: \[ x = \frac{7\pi}{24} + \frac{k\pi}{2} \quad \text{or} \quad x = \frac{11\pi}{24} + \frac{k\pi}{2} \] Thus, the general solution for \( x \) is: \[ x = \frac{7\pi}{24} + \frac{k\pi}{2} \quad \text{and} \quad x = \frac{11\pi}{24} + \frac{k\pi}{2} \quad (k \in \mathbb{Z}) \]