en: \( 4 \sin 2 \theta-3 \cos 2 \theta=0 \) Show that \( \tan 2 \theta=\frac{3}{4} \) Hence, solve for \( \theta \) where \( \theta \in\left(-360^{\circ} ;-90^{\circ}\right) \)

Let's start with the equation given: \[ 4 \sin 2\theta - 3 \cos 2\theta = 0 \] Rearranging this gives: \[ 4 \sin 2\theta = 3 \cos 2\theta \] Dividing both sides by \( \cos 2\theta \) (provided \( \cos 2\theta \neq 0 \)), we have: \[ \frac{4 \sin 2\theta}{\cos 2\theta} = 3 \] This simplifies to: \[ 4 \tan 2\theta = 3 \] Now, isolating \( \tan 2\theta \) results in: \[ \tan 2\theta = \frac{3}{4} \] Now, to solve for \( \theta \) in the interval \( (-360^\circ, -90^\circ) \): 1. The general solution for \( 2\theta \) can be expressed using the arctangent: \[ 2\theta = \tan^{-1}\left(\frac{3}{4}\right) + k \cdot 180^\circ \quad (k \in \mathbb{Z}) \] Calculating \( \tan^{-1}\left(\frac{3}{4}\right) \) gives: \[ 2\theta \approx 36.87^\circ \] So the first solution is: \[ 2\theta \approx 36.87^\circ + k \cdot 180^\circ \] 2. The value of \( \theta \) becomes: For \( k = 0 \): \[ 2\theta \approx 36.87^\circ \Rightarrow \theta \approx 18.44^\circ \] For \( k = 1 \): \[ 2\theta \approx 216.87^\circ \Rightarrow \theta \approx 108.44^\circ \] For \( k = -1 \): \[ 2\theta \approx -143.13^\circ \Rightarrow \theta \approx -71.56^\circ \] 3. The solutions that fit within the interval \( (-360^\circ, -90^\circ) \): - The only valid solution here is: \[ \theta \approx -71.56^\circ \] Therefore, the solution for \( \theta \) is approximately \( -71.56^\circ \) within the specified interval.