\( \frac{4 \cos (-x) \cdot \cos \left(90^{\circ}+x\right)}{\sin \left(30^{\circ}-1\right) \cdot \cos x+\cos \left(36^{\circ}-x\right) \cdot \sin x} \) 193 Detemine the general solution of \( \cos 2 x-7 \cos x-3=0 \).

To solve the equation \( \cos 2x - 7 \cos x - 3 = 0 \), we can use the double angle identity \( \cos 2x = 2 \cos^2 x - 1 \) to rewrite it as: \[ 2 \cos^2 x - 1 - 7 \cos x - 3 = 0 \] This simplifies to: \[ 2 \cos^2 x - 7 \cos x - 4 = 0 \] Letting \( y = \cos x \), we can then rewrite it as a quadratic equation: \[ 2y^2 - 7y - 4 = 0 \] Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ y = \frac{7 \pm \sqrt{(-7)^2 - 4(2)(-4)}}{2(2)} = \frac{7 \pm \sqrt{49 + 32}}{4} = \frac{7 \pm \sqrt{81}}{4} = \frac{7 \pm 9}{4} \] This gives us the two potential solutions for \( y \): \[ y_1 = \frac{16}{4} = 4 \quad \text{(not possible for } \cos x \text{)} \] \[ y_2 = \frac{-2}{4} = -\frac{1}{2} \] Thus, we only consider \( \cos x = -\frac{1}{2} \). The general solutions for \( \cos x = -\frac{1}{2} \) are: \[ x = 120^{\circ} + 360^{\circ}k \quad \text{and} \quad x = 240^{\circ} + 360^{\circ}k, \quad k \in \mathbb{Z} \] Or, in radians: \[ x = \frac{2\pi}{3} + 2\pi k \quad \text{and} \quad x = \frac{4\pi}{3} + 2\pi k \quad k \in \mathbb{Z} \] So, the complete general solution is given by these two sets of angles!